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Partial sums of length of terms in A081368 where A081368(1) is set to 0.
+20
1
0, 2, 5, 9, 14, 21, 28, 36, 45, 55, 66, 77, 90, 104, 119, 135, 152, 170
COMMENTS
Previous name was: The sequence of powers necessary to reconstruct Exp[0] from Thanh Diep's sequence A081368: E=Sum[ A081368[n]/10^a(n),{n,1,Length}].
REFERENCES
C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 350-351.
MATHEMATICA
a = {2, 71, 828, 1828, 45904, 5235360, 2874713, 52662497, 757247093, 6999595749, 66967627724, 76630353547, 5945713821785, 25166427427466, 391932003059921, 8174135966290435, 72900334295260595, 630738132328627943};
b = Table[Length[IntegerDigits[a[[n]]]], {n, 1, Length[a]}];
c = Table[Sum[b[[m]], {m, 1, n}] - 1, {n, 1, Length[b]}] Sum[a[[n]]/10^(c[[n]]), {n, 1, Length[a]}];
N[% - E, 100]
PROG
(PARI) v=[71, 828, 1828, 45904, 5235360, 2874713, 52662497, 757247093, 6999595749, 66967627724, 76630353547, 5945713821785, 25166427427466, 391932003059921, 8174135966290435, 72900334295260595, 630738132328627943];
concat([0], vector(#v, n, sum(j=1, n, #digits(v[j])))) \\ Joerg Arndt, Aug 13 2013
3, 14, 159, 2653, 58979, 323846, 2643383, 27950288, 419716939, 9375105820, 97494459230, 781640628620, 8998628034825, 34211706798214, 808651328230664, 7093844609550582, 23172535940812848, 111745028410270193, 8521105559644622948, 95493038196442881097
COMMENTS
More precisely: the integer resulting from reading the "next n digits of Pi" in base 10, so leading zeros cannot be directly seen, but easily be "reconstructed" from the fact that the term will have less than n digits although it is made from n digits of Pi. - M. F. Hasler, Jan 06 2023 It seems that all terms have at least one prime factor that does not appear in the combined list of prime factors of the preceding terms of the sequence. - Mario Cortés, Aug 20 2020 [Checked up to n=65. - Michel Marcus, Aug 21 2020]
FORMULA
a(n) = floor( Pi * 10^(n*(n+1)/2-1) ) mod (10^n). - Carl R. White, Aug 13 2010
EXAMPLE
a(3) = 159 because after the first (a(1) = 3) and the next two digits of Pi (a(2) = 14) the next three are 159.
Other examples are as follows and fall into a triangular digit pattern, though there is no guarantee that they will remain triangular in all cases
a(1) = 3;
a(2) = 14;
a(3) = 159;
a(4) = 2653;
a(5) = 58979;
(End)
Indeed, precisely whenever A086639(n) = 0, then the corresponding term of this sequence will lack one or more leading zeros and therefore the above list will deviate from the triangular shape. - M. F. Hasler, Jan 06 2023
MAPLE
Partitioner := proc(cons, len) local i, R, spl; R := []; i:=0;
spl := L -> [seq([seq(L[i], i=1 + n*(n+1)/2..(n+1)*(n+2)/2)], n=0..len)]:
ListTools:-Reverse(convert(floor(cons*10^((len+1)*(len+2)/2)), base, 10)):
map(`@`(parse, cat, op), spl(%)) end:
aList := -> Partitioner(Pi, 20); aList(20); # Peter Luschny, Aug 22 2020
MATHEMATICA
With[{pi=RealDigits[Pi, 10, 500][[1]]}, FromDigits/@Table[Take[pi, {n (n-1)/2+1, (n(n+1))/2}], {n, 25}]] (* Harvey P. Dale, Dec 24 2011 *)
PROG
(PARI) lista(nn) = {my(nd = 5+nn*(nn+1)/2); default(realprecision, nd); my(vd = digits(floor(Pi*10^nd))); my(pos = 1); my(vr = vector(nn)); for (n=1, nn, vr[n] = fromdigits(vector(n, k, vd[k+ pos-1])); pos += n; ); vr; } \\ Michel Marcus, Aug 21 2020
a(n) = the next n digits of phi, the golden ratio.
+10
3
1, 61, 803, 3988, 74989, 484820, 4586834, 36563811, 772030917, 9805762862, 13544862270, 526046281890, 2449707207204, 18939113748475, 408807538689175, 2126633862223536, 93179318006076672, 635443338908659593
MATHEMATICA
With[{phi=RealDigits[GoldenRatio, 10, 500][[1]]}, FromDigits/@Table[Take[ phi, {n (n-1)/2+1, (n(n+1))/2}], {n, 25}]] (* Harvey P. Dale, Dec 24 2011 *)
PROG
(PARI) { default(realprecision, 20180); x = (1 + sqrt(5))/2; for (n=1, 200, d=floor(x); x=(x-d)*10^(n+1); write("b093473.txt", n, " ", d)); } \\ Harry J. Smith, Jun 19 2009
a(0) = 3; for n > 0, break up decimal expansion of Pi into chunks of increasing lengths; leading zeros are not printed.
+10
2
3, 1, 41, 592, 6535, 89793, 238462, 6433832, 79502884, 197169399, 3751058209, 74944592307, 816406286208, 9986280348253, 42117067982148, 86513282306647, 938446095505822, 31725359408128481, 117450284102701938, 5211055596446229489
REFERENCES
Sylvia Nasar, A Beautiful Mind (1998), p. 210.
Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 64.
MAPLE
with(StringTools): lim:=23: s:=convert(evalf[lim^2](Pi-3), string): printf("3, "): for n from 1 to lim do printf("%d, ", parse(SubString(s, (n-1)*n/2+2..n*(n+1)/2+1))); od: # Nathaniel Johnston, May 08 2011
MATHEMATICA
Join[{3}, FromDigits/@With[{p=RealDigits[Pi, 10, 220][[1]]}, Table[ Take[ p, {(n(n-1))/2+2, (n(n-1))/2+1+n}], {n, 20}]]] (* Harvey P. Dale, Aug 20 2011 *)
Next n digits of sqrt(2).
+10
1
1, 41, 421, 3562, 37309, 504880, 1688724, 20969807, 856967187, 5376948073, 17667973799, 73247846210, 7038850387534, 32764157273501, 384623091229702, 4924836055850737, 21264412149709993, 583141322266592750, 5592755799950501152, 78206057147010955997
FORMULA
a(n) = floor(sqrt(2)*10^(n*(n + 1)/2 - 1)) mod (10^n).
EXAMPLE
a(2) = 41 because the second and third digits of sqrt(2) are 4 and 1.
MATHEMATICA
Table[Mod[Floor[Sqrt[2] 10^(n ((n + 1)/2) - 1)], 10^n], {n, 1, 20}]
Table[Floor[10^(-1 + (n (1 + n))/2) Sqrt[2]] + Ceiling[-(Floor[10^(-1 + (n (1 + n))/2) Sqrt[2]]/10^n)] 10^n, {n, 1, 20}]
With[{x=20}, FromDigits/@TakeList[RealDigits[Sqrt[2], 10, (x(x+1))/2] [[1]], Range[x]]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, May 04 2019 *)
PROG
(Magma) [Floor(Sqrt(2)*10^(n*(n + 1)/2 - 1)) mod (10^n): n in [1..30]]; // Vincenzo Librandi, Feb 15 2016 (PARI) a(n) = lift(Mod(floor(sqrt(2)*10^(n*(n + 1)/2 - 1)), 10^n)); \\ G. C. Greubel, Oct 07 2018
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