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Third partial sums of sixth powers ( A001014).
+20
23
1, 67, 927, 6677, 32942, 126378, 404634, 1129854, 2833479, 6515509, 13947505, 28115451, 53846156, 98669156, 173975076, 296541132, 490504893, 789878583, 1241708083, 1909993393, 2880500634, 4266609710, 6216356510, 8920844010, 12624212835, 17635378761
COMMENTS
This is one of a sequence of arrays that are the convolutions of the zero-padded sequences binomial(2n-1+k,k) with the Eulerian polynomials E(n,x) of A008292, represented by E(n,x) (1-x)^(-2n), which generate increasing partial sums of powers of integers: n= 2) (1 + 4*x + x^2)/(1-x)^4 is the o.g.f. of A000578, the convolution of (1,4,1) with A000292, giving the powers of m^3. n= 3) (1 + 11*x + 11*x^2 + x^3)/(1-x)^6 is the o.g.f. of A000538, convolution of (1,11,11,1) with A000389, giving the partial sums of m^4. n= 4) (1 + 26*x + 66*x^2 + 26*x^3 + x^4)/(1-x)^8, the o.g.f. of A101092, convolution of (1,26,66,26,1) with A000580, the second partial sums of m^5 n= 5) (1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1-x)^10, the o.g.f. of A254460, convolution of (1,57,302,302,57,1) with A000582, giving the third partial sums of m^6. - Tom Copeland, Dec 07 2015
LINKS
Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).
FORMULA
a(n) = n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2 -30*n +35*n^2 +30*n^3 +5*n^4)/5040.
G.f.: x*(1+x)*(1 +56*x +246*x^2 +56*x^3 +x^4)/(1-x)^10. - Colin Barker, Feb 04 2015
MAPLE
seq(binomial(n+3, 4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210, n=1..30); # G. C. Greubel, Aug 28 2019
MATHEMATICA
Table[n(1+n)(2+n)(3+n)(3+2n)(2 -30n +35n^2 +30n^3 +5n^4)/5040, {n, 30}] (* or *) CoefficientList[Series[(x+1)(x^4 +56x^3 +246x^2 +56x +1)/(x - 1)^10, {x, 0, 30}], x] (* Vincenzo Librandi, Feb 05 2015 *)
PROG
(PARI) vector(30, n, n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+5*n^4)/5040) \\ Colin Barker, Feb 04 2015 (Magma) [n*(1+n)*(2+n)*(3+n)*(3+2*n)*(2-30*n+35*n^2+30*n^3+ 5*n^4)/5040: n in [1..30]]; // Vincenzo Librandi, Feb 05 2015 (Sage) [binomial(n+3, 4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210 for n in (1..30)] # G. C. Greubel, Aug 28 2019 (GAP) List([1..30], n-> Binomial(n+3, 4)*(2*n+3)*(5*n^4 +30*n^3 +35*n^2 -30*n +2)/210); # G. C. Greubel, Aug 28 2019 (Python)
def A254640(n): return n*(n*(n*(n*(n*(n*(n*(n*(10*n + 135) + 720) + 1890) + 2394) + 945) - 640) - 450) + 36)//5040 # Chai Wah Wu, Dec 07 2021
Second partial sums of sixth powers ( A001014).
+20
9
1, 66, 860, 5750, 26265, 93436, 278256, 725220, 1703625, 3682030, 7431996, 14167946, 25730705, 44823000, 75305920, 122566056, 193963761, 299373690, 451829500, 668285310, 970507241, 1386109076, 1949746800, 2704487500
FORMULA
a(n) = n*(1 + n)^2*(2 + n)*(-1 + n*(2 + n))*(-2 + 3*n*(2 + n))/168.
G.f.: x*(1+x)*(1 + 56*x + 246*x^2 + 56*x^3 + x^4)/(1-x)^9. - Colin Barker, Dec 18 2012 a(n) = Sum_{i=1..n} i*(n+1-i)^6, by the definition. - Bruno Berselli, Jan 31 2014
MAPLE
f:=n->(3*n^8-14*n^6+21*n^4-10*n^2)/168;
MATHEMATICA
CoefficientList[Series[(x+1)(x^4+56x^3+246x^2+56x+1)/(1-x)^9, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 24 2014 *) Nest[Accumulate, Range[30]^6, 2] (* or *) LinearRecurrence[{9, -36, 84, -126, 126, -84, 36, -9, 1}, {1, 66, 860, 5750, 26265, 93436, 278256, 725220, 1703625}, 30] (* Harvey P. Dale, Jun 05 2019 *)
PROG
(Magma) [n*(1+n)^2*(2+n)*(-1+n*(2+n))*(-2+3*n*(2+n))/168: n in [1..40]]; // Vincenzo Librandi, Mar 24 2014 (PARI) vector(30, n, m=n+1; m^2*(3*m^6 -14*m^4 +21*m^2 -10)/168) \\ G. C. Greubel, Aug 28 2019 (Sage) [n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168 for n in (2..30)] # G. C. Greubel, Aug 28 2019 (GAP) List([2..30], n-> n^2*(3*n^6 -14*n^4 +21*n^2 -10)/168); # G. C. Greubel, Aug 28 2019
AUTHOR
Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 15 2004
Fourth partial sums of sixth powers ( A001014).
+20
7
1, 68, 995, 7672, 40614, 166992, 571626, 1701480, 4534959, 11050468, 24997973, 53113424, 106959580, 205628736, 379603812, 676144944, 1166649837, 1956528420, 3198236503, 5108229896, 7988730530, 12255340240
LINKS
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
FORMULA
G.f.: x*(1 + 57*x + 302*x^2 + 302*x^3 + 57*x^4 + x^5)/(1 - x)^11.
a(n) = n*(1 + n)*(2 + n)^2*(3 + n)*(4 + n)*(- 1 - 8*n + 14*n^2 + 8*n^3 + n^4)/5040.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + n^6.
EXAMPLE
First differences: 1, 63, 665, 3367, 11529, 31031, ... ( A022522) --------------------------------------------------------------------------
The sixth powers: 1, 64, 729, 4096, 15625, 46656, ... ( A001014) --------------------------------------------------------------------------
First partial sums: 1, 65, 794, 4890, 20515, 67171, ... ( A000540) Second partial sums: 1, 66, 860, 5750, 26265, 93436, ... ( A101093) Third partial sums: 1, 67, 927, 6677, 32942, 126378, ... ( A101099) Fourth partial sums: 1, 68, 995, 7672, 40614, 166992, ... (this sequence)
MAPLE
seq(binomial(n+4, 5)*(n+2)*((n^2+4*n-1)^2-2)/42, n=1..30); # G. C. Greubel, Aug 28 2019
MATHEMATICA
Table[n (1 + n) (2 + n)^2 (3 + n) (4 + n) (- 1 - 8 n + 14 n^2 + 8 n^3 + n^4)/5040, {n, 22}] (* or *)
Accumulate[Accumulate[Accumulate[Accumulate[Range[22]^6]]]] (* or *)
CoefficientList[Series[(- 1 - 57 x - 302 x^2 - 302 x^3 - 57 x^4 - x^5)/(- 1 + x)^11, {x, 0, 21}], x]
Nest[Accumulate, Range[30]^6, 4] (* or *) LinearRecurrence[{11, -55, 165, -330, 462, -462, 330, -165, 55, -11, 1}, {1, 68, 995, 7672, 40614, 166992, 571626, 1701480, 4534959, 11050468, 24997973}, 30] (* Harvey P. Dale, Dec 27 2015 *)
PROG
(PARI) vector(30, n, binomial(n+4, 5)*(n+2)*((n^2+4*n-1)^2-2)/42) \\ G. C. Greubel, Aug 28 2019 (Magma) [Binomial(n+4, 5)*(n+2)*((n^2+4*n-1)^2-2)/42: n in [1..30]]; // G. C. Greubel, Aug 28 2019 (Sage) [binomial(n+4, 5)*(n+2)*((n^2+4*n-1)^2-2)/42 for n in (1..30)] # G. C. Greubel, Aug 28 2019 (GAP) List([1..30], n-> Binomial(n+4, 5)*(n+2)*((n^2+4*n-1)^2-2)/42); # G. C. Greubel, Aug 28 2019
CROSSREFS
Cf. A254644 (fourth partial sums of fifth powers), A254646 (fourth partial sums of seventh powers).
Fifth partial sums of sixth powers ( A001014).
+20
7
1, 69, 1064, 8736, 49350, 216342, 787968, 2489448, 7024407, 18074875, 43072848, 96186272, 203145852, 408774588, 788378400, 1464523344, 2631173181, 4587701601, 7785938104, 12894168000, 20882898530, 33138238770
LINKS
Index entries for linear recurrences with constant coefficients, signature (12,-66,220,-495,792,-924,792,-495,220,-66,12,-1).
FORMULA
G.f.: (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)/(- 1 + x)^12.
a(n) = n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(5 + 2*n)*(- 3 + 5*n + n^2)*(4 + 15*n + 3*n^2)/332640.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + n^6.
EXAMPLE
First differences: 1, 63, 665, 3367, 11529, ... ( A022522) --------------------------------------------------------------------------
The sixth powers: 1, 64, 729, 4096, 15625, ... ( A001014) --------------------------------------------------------------------------
First partial sums: 1, 65, 794, 4890, 20515, ... ( A000540) Second partial sums: 1, 66, 860, 5750, 26265, ... ( A101093) Third partial sums: 1, 67, 927, 6677, 32942, ... ( A254640) Fourth partial sums: 1, 68, 995, 7672, 40614, ... ( A254645) Fifth partial sums: 1, 69, 1064, 8736, 49350, ... (this sequence)
MATHEMATICA
Table[n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (5 + 2*n) (- 3 + 5*n + n^2) (4 + 15 n + 3 n^2)/332640, {n, 22}] (* or *)
CoefficientList[Series[(1 + 57 x + 302 x^2 + 302 x^3 + 57 x^4 + x^5)/(- 1 + x)^12, {x, 0, 21}], x]
Sixth partial sums of sixth powers ( A001014).
+20
6
1, 70, 1134, 9870, 59220, 275562, 1063530, 3552978, 10577385, 28652260, 71725108, 167911380, 371057232, 779831820, 1568210220, 3032733564, 5663906745, 10251608346, 18037546450, 30931714450, 51814612980, 84952851750, 136562787270, 215565263550, 334584493425
LINKS
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).
FORMULA
G.f.: (-x - 57*x^2 - 302*x^3 - 302*x^4 - 57*x^5 - x^6)/(- 1 + x)^13.
a(n) = n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/665280.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + n^6.
Sum_{n>=1} 1/a(n) = 25622179/76545 - 3080*Pi^2/81 + 149600*Pi*tan(sqrt(37)*Pi/2)/(243*sqrt(37)). - Amiram Eldar, Jan 27 2022
EXAMPLE
First differences: 1, 63, 665, 3367, 11529, ... ( A022522) --------------------------------------------------------------------------
The sixth powers: 1, 64, 729, 4096, 15625, ... ( A001014) --------------------------------------------------------------------------
First partial sums: 1, 65, 794, 4890, 20515, ... ( A000540) Second partial sums: 1, 66, 860, 5750, 26265, ... ( A101093) Third partial sums: 1, 67, 927, 6677, 32942, ... ( A254640) Fourth partial sums: 1, 68, 995, 7672, 40614, ... ( A254645) Fifth partial sums: 1, 69, 1064, 8736, 49350, ... ( A254683) Sixth partial sums: 1, 70, 1134, 9870, 59220, ... (this sequence)
MATHEMATICA
Table[n (1 + n) (2 + n) (3 + n)^2 (4 + n) (5 + n) (6 + n) (- 3 + 5 n + n^2) (3 + 7 n + n^2)/665280, {n, 22}] (* or *) CoefficientList[Series[(- 1 - 57 x - 302 x^2 - 302 x^3 - 57 x^4 - x^5)/(- 1 + x)^13, {x, 0, 28}], x]
PROG
(Magma) [n*(1+n)*(2+n)*(3+n)^2*(4+n)*(5+n)*(6+n)*(-3+5*n+n^2)* (3+7*n+n^2)/665280: n in [1..30]]; // Vincenzo Librandi, Feb 15 2015 (PARI) vector(50, n, n*(1 + n)*(2 + n)*(3 + n)^2*(4 + n)*(5 + n)*(6 + n)*(-3 + 5*n + n^2)*(3 + 7*n + n^2)/665280) \\ Derek Orr, Feb 19 2015
Seventh partial sums of sixth powers ( A001014).
+20
5
1, 71, 1205, 11075, 70295, 345857, 1409387, 4962365, 15539750, 44192010, 115917118, 283828498, 654885730, 1434717550, 3002927770, 6035661334, 11699568079, 21951176425, 39988722875, 70920437325, 122735050305
LINKS
Index entries for linear recurrences with constant coefficients, signature (14,-91,364,-1001,2002,-3003,3432,-3003,2002,-1001,364,-91,14,-1).
FORMULA
G.f.: (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)/(- 1 + x)^14.
a(n) = (n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*(6 + n)*(7 + n)*(7 + 2*n)*(- 49 + 147*n^2 + 42*n^3 + 3*n^4))/51891840.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7) + n^6.
EXAMPLE
First differences: 1, 63, 665, 3367, 11529, ... ( A022522) --------------------------------------------------------------------
The sixth powers: 1, 64, 729, 4096, 15625, ... ( A001014) --------------------------------------------------------------------
First partial sums: 1, 65, 794, 4890, 20515, ... ( A000540) Second partial sums: 1, 66, 860, 5750, 26265, ... ( A101093) Third partial sums: 1, 67, 927, 6677, 32942, ... ( A254640) Fourth partial sums: 1, 68, 995, 7672, 40614, ... ( A254645) Fifth partial sums: 1, 69, 1064, 8736, 49350, ... ( A254683) Sixth partial sums: 1, 70, 1134, 9870, 59220, ... ( A254472) Seventh partial sums: 1, 71, 1205, 11075, 70295, ... (this sequence)
MATHEMATICA
Table[(n (1 + n) (2 + n) (3 + n) (4 + n) (5 + n) (6 + n) (7 + n) (7 + 2 n) (- 49 + 147 n^2 + 42 n^3 + 3 n^4))/51891840, {n, 21}] (* or *)
CoefficientList[Series[(1 + 57 x + 302 x^2 + 302 x^3 + 57 x^4 + x^5)/(- 1 + x)^14, {x, 0, 20}], x]
CROSSREFS
Cf. A000540, A001014, A022522, A101093, A254472, A254640, A254645, A254683, A254869, A254870, A254871.
Exponential transform of the sixth powers A001014.
+20
2
1, 1, 65, 922, 19685, 572036, 16379797, 542459296, 20028938953, 787480005520, 33447797179721, 1522102664036384, 73362723948758125, 3738119667151161280, 200625910519541044189, 11290451562860730241216, 664399657108812332697233, 40781390340823661046136064
FORMULA
E.g.f.: exp(exp(x)*(x^6+15*x^5+65*x^4+90*x^3+31*x^2+x)).
MAPLE
a:= proc(n) option remember; `if`(n=0, 1,
add(binomial(n-1, j-1)*j^6*a(n-j), j=1..n))
end:
seq(a(n), n=0..25);
Exponentially odd numbers.
+10
106
1, 2, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 26, 27, 29, 30, 31, 32, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 82, 83, 85, 86, 87, 88, 89, 91, 93, 94, 95, 96, 97
COMMENTS
The sequence is formed by 1 and the numbers whose prime power factorization contains only odd exponents.
The density of the sequence is the constant given by A065463. The term "exponentially odd integers" was apparently coined by Cohen (1960). These numbers were also called "unitarily 2-free", or "2-skew", by Cohen (1961). - Amiram Eldar, Jan 22 2024
FORMULA
Sum_{a(n)<=x} 1 = C*x + O(sqrt(x)*log x*e^(c*sqrt(log x)/(log(log x))), where c = 4*sqrt(2.4/log 2) = 7.44308... and C = Product_{prime p} (1 - 1/p*(p + 1)) = 0.7044422009991... ( A065463). Sum_{n>=1} 1/a(n)^s = zeta(2*s) * Product_{p prime} (1 + 1/p^s - 1/p^(2*s)), s>1. - Amiram Eldar, Sep 26 2023
MATHEMATICA
Select[Range@ 100, AllTrue[Last /@ FactorInteger@ #, OddQ] &] (* Version 10, or *)
Select[Range@ 100, Times @@ Boole[OddQ /@ Last /@ FactorInteger@ #] == 1 &] (* Michael De Vlieger, Feb 02 2016 *)
PROG
(PARI) isok(n)=my(f = factor(n)); for (k=1, #f~, if (!(f[k, 2] % 2), return (0))); 1; \\ Michel Marcus, Feb 02 2016 (Python)
from itertools import count, islice
from sympy import factorint
def A268335_gen(startvalue=1): # generator of terms >= startvalue return filter(lambda n:all(e&1 for e in factorint(n).values()), count(max(startvalue, 1)))
CROSSREFS
Cf. A002035, A209061, A138302, A197680, A000578, A000584, A001014, A001017, A008456, A010803, A010805, A010806, A010808, A010811, A010812, A001221, A124010.
a(n) = sigma_6(n), the sum of the 6th powers of the divisors of n.
+10
93
1, 65, 730, 4161, 15626, 47450, 117650, 266305, 532171, 1015690, 1771562, 3037530, 4826810, 7647250, 11406980, 17043521, 24137570, 34591115, 47045882, 65019786, 85884500, 115151530, 148035890, 194402650, 244156251, 313742650, 387952660, 489541650, 594823322, 741453700
COMMENTS
If the canonical factorization of n into prime powers is the product of p^e(p) then sigma_k(n) = Product_p ((p^((e(p)+1)*k))-1)/(p^k-1).
Sum_{d|n} 1/d^k is equal to sigma_k(n)/n^k. So sequences A017665- A017712 also give the numerators and denominators of sigma_k(n)/n^k for k = 1..24. The power sums sigma_k(n) are in sequences A000203 (k=1), A001157- A001160 (k=2,3,4,5), A013954- A013972 for k = 6,7,...,24. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 05 2001
FORMULA
L.g.f.: -log(Product_{k>=1} (1 - x^k)^(k^5)) = Sum_{n>=1} a(n)*x^n/n. - Ilya Gutkovskiy, May 06 2017 Multiplicative with a(p^e) = (p^(6*e+6)-1)/(p^6-1).
Dirichlet g.f.: zeta(s)*zeta(s-6).
Sum_{k=1..n} a(k) = zeta(7) * n^7 / 7 + O(n^8). (End)
PROG
(Magma) [DivisorSigma(6, n): n in [1..30]]; // Bruno Berselli, Apr 10 2013
Square triangular numbers: numbers that are both triangular and square.
(Formerly M5259 N2291)
+10
83
0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, 63955431761796, 2172602007770041, 73804512832419600, 2507180834294496361, 85170343853180456676, 2893284510173841030625, 98286503002057414584576, 3338847817559778254844961, 113422539294030403250144100
COMMENTS
Satisfies a recurrence of S_r type for r=36: 0, 1, 36 and a(n-1)*a(n+1)=(a(n)-1)^2. First observed by Colin Dickson in alt.math.recreational, Mar 07 2004. - Rainer Rosenthal, Mar 14 2004 For every n, a(n) is the first of three triangular numbers in geometric progression. The third number in the progression is a(n+1). The middle triangular number is sqrt(a(n)*a(n+1)). Chen and Fang prove that four distinct triangular numbers are never in geometric progression. - T. D. Noe, Apr 30 2007 Conjecture: No a(2^k), where k is a nonnegative integer, can be expressed as a sum of a positive square number and a positive triangular number. - Ivan N. Ianakiev, Sep 19 2012 For n > 0, these are the triangular numbers which are the sum of two consecutive triangular numbers, for instance 36 = 15 + 21 and 1225 = 595 + 630. - Michel Marcus, Feb 18 2014 The sequence is the case P1 = 36, P2 = 68, Q = 1 of the 3-parameter family of 4th order linear divisibility sequences found by Williams and Guy. - Peter Bala, Apr 03 2014 For n=2k, k > 0, a(n) is divisible by 12 and is therefore abundant. I conjecture that for n=2k+1 a(n) is deficient [true for k up to 43 incl.]. - Ivan N. Ianakiev, Sep 30 2014 The conjecture is true for all k > 0 because: For n=2k+1, k > 0, a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. - Muniru A Asiru, Apr 13 2016
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 193.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923; see Vol. 2, p. 10.
Martin Gardner, Time Travel and other Mathematical Bewilderments, Freeman & Co., 1988, pp. 16-17.
Miodrag S. Petković, Famous Puzzles of Great Mathematicians, Amer. Math. Soc. (AMS), 2009, p. 64.
J. H. Silverman, A Friendly Introduction to Number Theory, Prentice Hall, 2001, p. 196.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 93.
LINKS
Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7. J. L. Pietenpol, A. V. Sylwester, E. Just, and R. M. Warten, Problem E 1473, Amer. Math. Monthly, Vol. 69, No. 2 (Feb. 1962), pp. 168-169. (From the editorial note on p. 169 of this source, we learn that the question about the existence of perfect squares in the sequence of triangular numbers cropped up in the Euler-Goldbach Briefwechsel of 1730; the translation into English of the relevant letters can be found at Correspondence of Leonhard Euler with Christian Goldbach (part II), pp. 614-615.) - José Hernández, May 24 2022 Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
FORMULA
a(0) = 0, a(1) = 1; for n >= 2, a(n) = 34 * a(n-1) - a(n-2) + 2.
G.f.: x*(1 + x) / (( 1 - x )*( 1 - 34*x + x^2 )).
a(n-1) * a(n+1) = (a(n)-1)^2. - Colin Dickson, posting to alt.math.recreational, Mar 07 2004
If L is a square-triangular number, then the next one is 1 + 17*L + 6*sqrt(L + 8*L^2). - Lekraj Beedassy, Jun 27 2001 a(n) - a(n-1) = A046176(n). - Sophie Kuo (ejiqj_6(AT)yahoo.com.tw), May 27 2006 a(n) = (((17+12*sqrt(2))^n) + ((17-12*sqrt(2))^n)-2)/32. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 26 2002
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2). See UWC problem link and solution. - Jaap Spies, Dec 12 2004 a(n) = 35*(a(n-1) - a(n-2)) + a(n-3);
a(n) = -1/16 + ((-24 + 17*sqrt(2))/2^(11/2))*(17 - 12*sqrt(2))^(n-1) + ((24 + 17*sqrt(2))/2^(11/2))*(17 + 12*sqrt(2))^(n-1). (End)
Closed form (as square = triangular): ( (sqrt(2)+1)^(2*n)/(4*sqrt(2)) - (1-sqrt(2))^(2*n)/(4*sqrt(2)) )^2 = (1/2) * ( ( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2 + 1 )*( (sqrt(2)+1)^n / 2 - (sqrt(2)-1)^n / 2 )^2. - Bill Gosper, Jul 25 2008 a(n) = floor((17 + 12*sqrt(2))*a(n-1)) + 3 = floor(3*sqrt(2)/4 + (17 + 12*sqrt(2))*a(n-1) + 1). - Manuel Valdivia, Aug 15 2011 a(2*n) = A001333(2*n)^2 * ( A001333(2*n)^2 - 1)/2, and a(2*n+1) = A001333(2*n+1)^2 * ( A001333(2*n+1)^2 + 1)/2. The latter is equivalent to the comment above from Ivan using A002315, which is a bisection of A001333. Using A001333 shows symmetry and helps show that a(n) are both "squares of triangular" and "triangular of squares". - Richard R. Forberg, Aug 30 2013 a(n) = (T(n,17) - 1)/16, where T(n,x) denotes the Chebyshev polynomial of the first kind.
a(n) = U(n-1,3)^2, for n >= 1, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = the bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -17; 1, 18].
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End) a(n) = A000129(n)^4 + Sum_{k=0..( A000129(n)^2 - ( A000129(n) mod 2))} 2*k. This formula can be proved graphically by taking the corresponding triangle of a square triangular number and cutting both acute angles, one level at a time (sum of consecutive even numbers), resulting in a square of squares (4th powers). a(n) = A002965(2*n)^4 + Sum_{k= A002965(2*n)^2.. A002965(2*n)* A002965(2*n + 1) - 1} 2*k + 1. This formula takes an equivalent sum of consecutives, but odd numbers. (End) E.g.f.: (exp((17-12*sqrt(2))*x) + exp((17+12*sqrt(2))*x) - 2*exp(x))/32. - Ilya Gutkovskiy, Jul 16 2016
EXAMPLE
a(2) = ((17 + 12*sqrt(2))^2 + (17 - 12*sqrt(2))^2 - 2)/32 = (289 + 24*sqrt(2) + 288 + 289 - 24*sqrt(2) + 288 - 2)/32 = (578 + 576 - 2)/32 = 1152/32 = 36 and 6^2 = 36 = 8*9/2 => a(2) is both the 6th square and the 8th triangular number.
MAPLE
a:=17+12*sqrt(2); b:=17-12*sqrt(2); A001110:=n -> expand((a^n + b^n - 2)/32); seq( A001110(n), n=0..20); # Jaap Spies, Dec 12 2004
MATHEMATICA
Table[(1/8) Round[N[Sinh[2 n ArcSinh[1]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *) Transpose[NestList[Flatten[{Rest[#], 34Last[#]-First[#]+2}]&, {0, 1}, 20]][[1]] (* Harvey P. Dale, Mar 25 2011 *) LinearRecurrence[{35, -35, 1}, {0, 1, 36}, 20] (* T. D. Noe, Mar 25 2011 *) LinearRecurrence[{6, -1}, {0, 1}, 20]^2 (* Harvey P. Dale, Oct 22 2012 *) (* Square = Triangular = Triangular = A001110 *) ChebyshevU[#-1, 3]^2==Binomial[ChebyshevT[#/2, 3]^2, 2]==Binomial[(1+ChebyshevT[#, 3])/2, 2]=={1, 36, 1225, 41616, 1413721}[[#]]&@Range[5]
L=0; r={}; Do[AppendTo[r, L]; L=1+17*L+6*Sqrt[L+8*L^2], {i, 1, 19}]; r (* Kebbaj Mohamed Reda, Aug 02 2023 *)
PROG
(PARI) a=vector(100); a[1]=1; a[2]=36; for(n=3, #a, a[n]=34*a[n-1]-a[n-2]+2); a \\ Charles R Greathouse IV, Jul 25 2011 (Haskell)
a001110 n = a001110_list !! n
a001110_list = 0 : 1 : (map (+ 2) $
zipWith (-) (map (* 34) (tail a001110_list)) a001110_list)
(MIT/GNU Scheme, with memoizing definec-macro from Antti Karttunen's IntSeq-library) ;; The following two are for testing whether n is in this sequence:
(define (inA001110? n) (and (zero? ( A068527 n)) (inA001109? (floor->exact (sqrt n))))) (define (inA001109? n) (= (* 8 n n) (floor->exact (* (sqrt 8) n (ceiling->exact (* (sqrt 8) n))))))
(Magma) [n le 2 select n-1 else Floor((6*Sqrt(Self(n-1)) - Sqrt(Self(n-2)))^2): n in [1..20]]; // Vincenzo Librandi, Jul 22 2015
CROSSREFS
Cf. A240129 (triangular numbers that are squares of triangular numbers), A100047.
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