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Euler transform of A000292.
(Formerly M3859 N1581)
+20
35
1, 5, 15, 45, 120, 331, 855, 2214, 5545, 13741, 33362, 80091, 189339, 442799, 1023192, 2340904, 5302061, 11902618, 26488454, 58479965, 128120214, 278680698, 602009786, 1292027222, 2755684669, 5842618668, 12317175320, 25825429276, 53865355154, 111786084504, 230867856903, 474585792077, 971209629993
REFERENCES
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).
FORMULA
a(n) ~ Zeta(5)^(379/3600) / (2^(521/1800) * sqrt(5*Pi) * n^(2179/3600)) * exp(Zeta'(-1)/3 - Zeta(3) / (8*Pi^2) - Pi^16 / (3110400000 * Zeta(5)^3) + Pi^8*Zeta(3) / (216000 * Zeta(5)^2) - Zeta(3)^2 / (90*Zeta(5)) + Zeta'(-3)/6 + (Pi^12 / (10800000 * 2^(2/5) * Zeta(5)^(11/5)) - Pi^4 * Zeta(3) / (900 * 2^(2/5) * Zeta(5)^(6/5))) * n^(1/5) + (Zeta(3) / (3 * 2^(4/5) * Zeta(5)^(2/5)) - Pi^8 / (36000 * 2^(4/5) * Zeta(5)^(7/5))) * n^(2/5) + Pi^4 / (180 * 2^(1/5) * Zeta(5)^(3/5)) * n^(3/5) + 5*Zeta(5)^(1/5) / 2^(8/5) * n^(4/5)). - Vaclav Kotesovec, Mar 12 2015
MAPLE
with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d, j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:= etr(n-> binomial(n+2, 3)): seq(a(n), n=1..26); # Alois P. Heinz, Sep 08 2008
MATHEMATICA
max = 33; f[x_] := Exp[ Sum[ x^k/(1-x^k)^4/k, {k, 1, max}]]; Drop[ CoefficientList[ Series[ f[x], {x, 0, max}], x], 1](* Jean-François Alcover, Nov 21 2011, after Joerg Arndt *) nmax=50; Rest[CoefficientList[Series[Product[1/(1-x^k)^(k*(k+1)*(k+2)/6), {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Mar 11 2015 *) etr[p_] := Module[{b}, b[n_] := b[n] = If[n==0, 1, Sum[DivisorSum[j, #*p[#] &]*b[n-j], {j, 1, n}]/n]; b]; a = etr[Binomial[#+2, 3]&]; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Nov 24 2015, after Alois P. Heinz *)
PROG
(PARI) a(n)=if(n<1, 0, polcoeff(exp(sum(k=1, n, x^k/(1-x^k)^4/k, x*O(x^n))), n)) /* Joerg Arndt, Apr 16 2010 */ (PARI) N=66; x='x+O('x^66); gf=-1 + exp(sum(k=1, N, x^k/(1-x^k)^4/k)); Vec(gf) /* Joerg Arndt, Jul 06 2011 */ (Sage) # uses[EulerTransform from A166861] and prepends a(0) = 1. a = EulerTransform(lambda n: n*(n+1)*(n+2)//6)
Minimal number of tetrahedral numbers ( A000292(k) = k(k+1)(k+2)/6) needed to sum to n.
+20
35
1, 2, 3, 1, 2, 3, 4, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 2, 3, 4, 5, 3, 1, 2, 3, 4, 2, 2, 3, 4, 3, 3, 2, 3, 4, 4, 3, 3, 4, 5, 4, 4, 2, 1, 2, 3, 3, 2, 3, 4, 4, 3, 3, 2, 3, 4, 4, 2, 3, 4, 5, 3, 3, 2, 3, 4, 4, 3, 4, 5, 5, 1, 2, 3, 4, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 3, 4, 4, 3, 4
COMMENTS
According to Dickson, Pollock conjectures that a(n) <= 5 for all n. Watson shows that a(n) <= 8 for all n, and Salzer and Levine show that a(n) <= 5 for n <= 452479659. - N. J. A. Sloane, Jul 15 2011 Possible correction of the first comment by Sloane 2011: it appears to me from the linked reference by Salzer and Levine 1968 that 452479659 is instead the upper limit for sums of five Qx = Tx + x, where Tx are the tetrahedral numbers we want. They also mention an upper limit for sums of five Tx, which is: a(n) <= 5 for n <= 276976383. - Ewoud Dronkert, May 30 2024 Could be extended with a(0) = 0, in analogy to A061336. Kim (2003, first row of table "d = 3" on p. 73) gives max {a(n)} = 5 as a "numerical result", but the value has no "* denoting exact values" (see Remark at end of paper), which means this could be incorrect. - M. F. Hasler, Mar 06 2017, edited Sep 22 2022
REFERENCES
Dickson, L. E., History of the Theory of Numbers, Vol. 2: Diophantine Analysis. New York: Dover, 1952, see p. 13.
MAPLE
tet:=[seq((n^3-n)/6, n=1..20)];
LAGRANGE(tet, 8, 120); # the LAGRANGE transform of a sequence is defined in A193101. - N. J. A. Sloane, Jul 15 2011
PROG
(PARI)
seq(N) = {
my(a = vector(N, k, 8), T = k->(k*(k+1)*(k+2))\6);
for (n = 1, N,
my (k1 = sqrtnint((6*n)\8, 3), k2 = sqrtnint(6*n, 3));
while(n < T(k2), k2--); if (n == T(k2), a[n] = 1; next());
for (k = k1, k2, a[n] = min(a[n], a[n - T(k)] + 1))); a;
};
CROSSREFS
Cf. A000292 (tetrahedral numbers), A000797 (numbers that need 5 tetrahedral numbers). Cf. A061336 (analog for triangular numbers).
Number of compositions (ordered partitions) of n into tetrahedral (or triangular pyramidal) numbers ( A000292).
+20
16
1, 1, 1, 1, 2, 3, 4, 5, 7, 10, 15, 21, 29, 40, 57, 81, 114, 159, 223, 314, 444, 625, 878, 1233, 1736, 2445, 3441, 4838, 6804, 9573, 13473, 18957, 26668, 37514, 52780, 74264, 104488, 147000, 206808, 290961, 409369, 575955, 810314, 1140029, 1603924, 2256603, 3174867, 4466763, 6284339, 8841533, 12439323
FORMULA
G.f.: 1/(1 - Sum_{k>=1} x^(k*(k+1)*(k+2)/6)).
EXAMPLE
a(8) = 7 because we have [4, 4], [4, 1, 1, 1, 1], [1, 4, 1, 1, 1], [1, 1, 4, 1, 1], [1, 1, 1, 4, 1], [1, 1, 1, 1, 4] and [1, 1, 1, 1, 1, 1, 1, 1].
MATHEMATICA
nmax = 50; CoefficientList[Series[1/(1 - Sum[x^(k (k + 1) (k + 2)/6), {k, 1, nmax}]), {x, 0, nmax}], x]
PROG
(PARI) Vec(1/(1 - sum(k=1, 50, x^(k*(k + 1)*(k + 2)/6)) + O(x^51))) \\ Indranil Ghosh, Mar 15 2017
1, 4, 6, 8, 10, 12, 19, 20, 27, 35, 44, 48, 56, 64, 84, 85, 120, 124, 125, 146, 165, 216, 220, 231, 255, 286, 343, 344, 364, 455, 456, 489, 512, 560, 670, 680, 729, 742, 816, 891, 969, 1000, 1128, 1140, 1156, 1330, 1331, 1469, 1540, 1629, 1728, 1771, 1834
MATHEMATICA
nn = 25; t1 = Table[n (n + 1) (n + 2)/6, {n, nn}]; t2 = Table[n^3, {n, nn}]; t3 = Table[(2*n^3 + n)/3, {n, nn}]; t4 = Table[n (3*n - 1) (3*n - 2)/2, {n, nn}]; t5 = Table[n (5*n^2 - 5*n + 2)/2, {n, nn}]; Select[Union[t1, t2, t3, t4, t5], # <= t1[[-1]] &] (* T. D. Noe, Oct 13 2012 *)
PROG
(Haskell)
a053012 n = a053012_list !! (n-1)
a053012_list = tail $ f
[a000292_list, a000578_list, a005900_list, a006566_list, a006564_list]
where f pss = m : f (map (dropWhile (<= m)) pss)
where m = minimum (map head pss)
(PARI) listpoly(lim, poly[..])=my(v=List()); for(i=1, #poly, my(P=poly[i], x=variable(P), f=k->subst(P, x, k), n, t); while((t=f(n++))<=lim, listput(v, t))); Set(v)
list(lim)=my(n='n); listpoly(lim, n*(n+1)*(n+2)/6, n^3, (2*n^3+n)/3, n*(3*n-1)*(3*n-2)/2, n*(5*n^2-5*n+2)/2) \\ Charles R Greathouse IV, Oct 11 2016
AUTHOR
Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Feb 22 2000
Number T(n,k) of compositions of n into k parts with distinct multiplicities, where parts are counted without multiplicities; triangle T(n,k), n>=0, 0<=k<=max{i: A000292(i)<=n}, read by rows.
+20
10
1, 0, 1, 0, 2, 0, 2, 0, 3, 3, 0, 2, 10, 0, 4, 12, 0, 2, 38, 0, 4, 56, 0, 3, 79, 0, 4, 152, 60, 0, 2, 251, 285, 0, 6, 284, 498, 0, 2, 594, 1438, 0, 4, 920, 2816, 0, 4, 1108, 5208, 0, 5, 2136, 11195, 0, 2, 3402, 24094, 0, 6, 4407, 38523, 0, 2, 8350, 85182
EXAMPLE
T(5,1) = 2: [1,1,1,1,1], [5].
T(5,2) = 10: [1,1,1,2], [1,1,2,1], [1,2,1,1], [2,1,1,1], [1,2,2], [2,1,2], [2,2,1], [1,1,3], [1,3,1], [3,1,1].
Triangle T(n,k) begins:
1;
0, 1;
0, 2;
0, 2;
0, 3, 3;
0, 2, 10;
0, 4, 12;
0, 2, 38;
0, 4, 56;
0, 3, 79;
0, 4, 152, 60;
MAPLE
b:= proc(n, i, s) option remember; `if`(n=0, add(j, j=s)!,
`if`(i<1, 0, expand(add(`if`(j>0 and j in s, 0, `if`(j=0, 1, x)*
b(n-i*j, i-1, `if`(j=0, s, s union {j}))/j!), j=0..n/i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n$2, {})):
seq(T(n), n=0..16);
MATHEMATICA
b[n_, i_, s_List] := b[n, i, s] = If[n == 0, Total[s]!, If[i<1, 0, Expand[ Sum[ If[j>0 && MemberQ[s, j], 0, If[j == 0, 1, x]*b[n-i*j, i-1, If[j == 0, s, s ~Union~ {j}]]/j!], {j, 0, n/i}]]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n, {}]]; Table[T[n], {n, 0, 16}] // Flatten (* Jean-François Alcover, Feb 11 2015, after Alois P. Heinz *)
Number of partitions of n into exactly k different parts with distinct multiplicities; triangle T(n,k), n>=0, 0<=k<=max{i: A000292(i)<=n}, read by rows.
+20
9
1, 0, 1, 0, 2, 0, 2, 0, 3, 1, 0, 2, 3, 0, 4, 3, 0, 2, 8, 0, 4, 9, 0, 3, 12, 0, 4, 16, 1, 0, 2, 22, 4, 0, 6, 20, 5, 0, 2, 31, 12, 0, 4, 35, 16, 0, 4, 34, 24, 0, 5, 44, 33, 0, 2, 51, 52, 0, 6, 53, 57, 0, 2, 62, 89, 0, 6, 65, 100, 1, 0, 4, 68, 131, 5, 0, 4, 87
EXAMPLE
T(0,0) = 1: [].
T(1,1) = 1: [1].
T(2,1) = 2: [1,1], [2].
T(4,1) = 3: [1,1,1,1], [2,2], [4].
T(4,2) = 1: [2,1,1]; part 2 occurs once and part 1 occurs twice.
T(5,2) = 3: [2,1,1,1], [2,2,1], [3,1,1].
T(7,2) = 8: [2,1,1,1,1,1], [2,2,1,1,1], [2,2,2,1], [3,1,1,1,1], [3,2,2], [3,3,1], [4,1,1,1], [5,1,1].
T(10,1) = 4: [1,1,1,1,1,1,1,1,1,1], [2,2,2,2,2], [5,5], [10].
T(10,3) = 1: [3,2,2,1,1,1].
Triangle T(n,k) begins:
1;
0, 1;
0, 2;
0, 2;
0, 3, 1;
0, 2, 3;
0, 4, 3;
0, 2, 8;
0, 4, 9;
0, 3, 12;
0, 4, 16, 1;
MAPLE
b:= proc(n, i, t, s) option remember;
`if`(nops(s)>t, 0, `if`(n=0, 1, `if`(i<1, 0, b(n, i-1, t, s)+
add(`if`(j in s, 0, b(n-i*j, i-1, t, s union {j})), j=1..n/i))))
end:
g:= proc(n) local i; for i while i*(i+1)*(i+2)/6<=n do od; i-1 end:
T:= n-> seq(b(n, n, k, {}) -b(n, n, k-1, {}), k=0..g(n)):
seq(T(n), n=0..30);
MATHEMATICA
b[n_, i_, t_, s_] := b[n, i, t, s] = If[Length[s] > t, 0, If[n == 0, 1, If[i < 1, 0, b[n, i-1, t, s] + Sum[If[MemberQ[s, j], 0, b[n-i*j, i-1, t, s ~Union~ {j}]], {j, 1, n/i}]]]]; g[n_] := Module[{i}, For[ i = 1, i*(i+1)*(i+2)/6 <= n , i++]; i-1 ]; t[n_] := Table [b[n, n, k, {}] - b[n, n, k-1, {}], {k, 0, g[n]}]; Table [t[n], {n, 0, 30}] // Flatten (* Jean-François Alcover, Dec 19 2013, translated from Maple *)
CROSSREFS
First row with length (t+1): A000292(t). Cf. A242896 (the same for compositions):
Exponential transform of binomial(n,3) = A000292(n-2).
+20
8
1, 0, 0, 1, 4, 10, 30, 175, 1176, 7084, 42120, 286605, 2270180, 19213766, 166326524, 1497096055, 14374680880, 147259920760, 1582837679056, 17659771122969, 204674606377140, 2473357218561250, 31148510170120420, 407154732691440811, 5504706823227724904
COMMENTS
a(n) is the number of ways of placing n labeled balls into indistinguishable boxes, where in each filled box 3 balls are seen at the top.
a(n) is also the number of forests of labeled rooted trees of height at most 1, with n labels, where each root contains 3 labels.
FORMULA
E.g.f.: exp(exp(x)*x^3/3!).
MAPLE
a:= proc(n) option remember; `if`(n=0, 1,
add(binomial(n-1, j-1) *binomial(j, 3) *a(n-j), j=1..n))
end:
seq(a(n), n=0..30);
MATHEMATICA
Table[Sum[BellY[n, k, Binomial[Range[n], 3]], {k, 0, n}], {n, 0, 25}] (* Vladimir Reshetnikov, Nov 09 2016 *)
Triplicated tetrahedral numbers A000292.
+20
8
1, 1, 1, 4, 4, 4, 10, 10, 10, 20, 20, 20, 35, 35, 35, 56, 56, 56, 84, 84, 84, 120, 120, 120, 165, 165, 165, 220, 220, 220, 286, 286, 286, 364, 364, 364, 455, 455, 455, 560, 560, 560, 680, 680, 680, 816, 816, 816, 969, 969, 969
COMMENTS
The Ca1 and Ze3 triangle sums, see A180662 for their definitions, of the triangle A159797 are linear sums of shifted versions of the triplicated tetrahedral numbers, e.g. Ca1(n) = a(n-1) + a(n-2) + 2*a(n-3) + a(n-6). The Ca1, Ca2, Ze3 and Ze4 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above.
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,0,3,-3,0,-3,3,0,1,-1).
FORMULA
a(n) = binomial(floor(n/3)+3,3).
a(n) + a(n-1) + a(n-2) = A144677(n). G.f.: 1/((x-1)^4*(x^2+x+1)^3).
MATHEMATICA
LinearRecurrence[{1, 0, 3, -3, 0, -3, 3, 0, 1, -1}, {1, 1, 1, 4, 4, 4, 10, 10, 10, 20}, 60] (* Harvey P. Dale, Mar 09 2018 *)
Quadruplicated tetrahedral numbers A000292.
+20
8
1, 1, 1, 1, 4, 4, 4, 4, 10, 10, 10, 10, 20, 20, 20, 20, 35, 35, 35, 35, 56, 56, 56, 56, 84, 84, 84, 84, 120, 120, 120, 120, 165, 165, 165, 165, 220, 220, 220, 220, 286, 286, 286, 286, 364, 364, 364, 364, 455, 455, 455, 455
COMMENTS
The Gi1 triangle sums, for the definitions of these and other triangle sums see A180662, of the triangle A159797 are linear sums of shifted versions of the quadruplicated tetrahedral numbers A000292, i.e., Gi1(n) = a(n-1) + a(n-2) + a(n-3) + 2*a(n-4) + a(n-8). The Gi1 and Gi2 triangle sums of the Connell sequence A001614 as a triangle are also linear sums of shifted versions of the sequence given above.
LINKS
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3,-3,0,0,-3,3,0,0,1,-1).
FORMULA
a(n) = binomial(floor(n/4)+3,3).
a(n-3) + a(n-2) + a(n-1) + a(n) = A144678(n). a(n) = +a(n-1) +3*a(n-4) -3*a(n-5) -3*a(n-8) +3*a(n-9) +a(n-12) -a(n-13).
G.f.: 1 / ( (1+x)^3*(1+x^2)^3*(x-1)^4 ).
MAPLE
A190718:= proc(n) binomial(floor(n/4)+3, 3) end:
MATHEMATICA
LinearRecurrence[{1, 0, 0, 3, -3, 0, 0, -3, 3, 0, 0, 1, -1}, {1, 1, 1, 1, 4, 4, 4, 4, 10, 10, 10, 10, 20}, 60] (* Harvey P. Dale, Oct 20 2012 *)
Numerator of sum of reciprocals of first n tetrahedral numbers A000292.
+20
7
1, 5, 27, 7, 10, 81, 35, 22, 81, 65, 77, 135, 52, 119, 405, 76, 85, 567, 209, 115, 378, 275, 299, 486, 175, 377, 1215, 217, 232, 1485, 527, 280, 891, 629, 665, 1053, 370, 779, 2457, 430, 451, 2835, 989, 517, 1620, 1127, 1175, 1836, 637, 1325, 4131, 715, 742
COMMENTS
Denominators are A118392. Fractions are: 1/1, 5/4, 27/20, 7/5, 10/7, 81/56, 35/24, 22/15, 81/55, 65/44, 77/52, 135/91, 52/35, 119/80, 405/272, 76/51, 85/57, 567/380, 209/140, 115/77, 378/253, 275/184, 299/200, 486/325, 175/117, 377/252, 1215/812, 217/145, 232/155, 1485/992. 2n+3 divides a(2n). 2n-1 divides a(2n-1). p divides a(p) for prime p>2. The only primes in a(n) are a(2) = 5 and a(4) = 7. - Alexander Adamchuk, May 08 2007
EXAMPLE
a(1) = 1 = numerator of 1/1.
a(2) = 5 = numerator of 5/4 = 1/1 + 1/4.
a(3) = 27 = numerator of 27/20 = 1/1 + 1/4 + 1/10.
a(4) = 7 = numerator of 7/5 = 1/1 + 1/4 + 1/10 + 1/20.
a(5) = 10 = numerator of 10/7 = 1/1 + 1/4 + 1/10 + 1/20 + 1/35.
a(20) = 115 = numerator of 115/77 = 1/1 + 1/4 + 1/10 + 1/20 + 1/35 + 1/56 + 1/84 + 1/120 + 1/165 + 1/220 + 1/286 + 1/364 + 1/455 + 1/560 + 1/680 + 1/816 + 1/969 + 1/1140 + 1/1330 + 1/1540.
MATHEMATICA
Table[ Numerator[3n(n+3)/(2(n+1)(n+2))], {n, 1, 100} ] (* Alexander Adamchuk, May 08 2007 *) Accumulate[1/Binomial[Range[60]+2, 3]]//Numerator (* Harvey P. Dale, Aug 31 2023 *)
PROG
(PARI) s=0; for(i=3, 50, s+=1/binomial(i, 3); print(numerator(s))) /* Phil Carmody, Mar 27 2012 */ (Sage) [numerator(3*n*(n+3)/(2*(n+1)*(n+2))) for n in (1..60)] # G. C. Greubel, Feb 18 2021 (Magma) [Numerator(3*n*(n+3)/(2*(n+1)*(n+2))): n in [1..60]]; // G. C. Greubel, Feb 18 2021
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