A152888
Partial sums of length of terms in A081368 where A081368(1) is set to 0.
Original entry on oeis.org
0, 2, 5, 9, 14, 21, 28, 36, 45, 55, 66, 77, 90, 104, 119, 135, 152, 170
Offset: 1
- C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 350-351.
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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]
-
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
A090897
Next n digits of Pi.
Original entry on oeis.org
3, 14, 159, 2653, 58979, 323846, 2643383, 27950288, 419716939, 9375105820, 97494459230, 781640628620, 8998628034825, 34211706798214, 808651328230664, 7093844609550582, 23172535940812848, 111745028410270193, 8521105559644622948, 95493038196442881097
Offset: 1
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.
From _Aaron T Cowan_, Jan 03 2023: (Start)
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
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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
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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 *)
-
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
A093473
a(n) = the next n digits of phi, the golden ratio.
Original entry on oeis.org
1, 61, 803, 3988, 74989, 484820, 4586834, 36563811, 772030917, 9805762862, 13544862270, 526046281890, 2449707207204, 18939113748475, 408807538689175, 2126633862223536, 93179318006076672, 635443338908659593
Offset: 1
-
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 *)
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{ 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
A136517
a(0) = 3; for n > 0, break up decimal expansion of Pi into chunks of increasing lengths; leading zeros are not printed.
Original entry on oeis.org
3, 1, 41, 592, 6535, 89793, 238462, 6433832, 79502884, 197169399, 3751058209, 74944592307, 816406286208, 9986280348253, 42117067982148, 86513282306647, 938446095505822, 31725359408128481, 117450284102701938, 5211055596446229489
Offset: 0
- Sylvia Nasar, A Beautiful Mind (1998), p. 210.
- Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 64.
-
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
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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 *)
A267325
Next n digits of sqrt(2).
Original entry on oeis.org
1, 41, 421, 3562, 37309, 504880, 1688724, 20969807, 856967187, 5376948073, 17667973799, 73247846210, 7038850387534, 32764157273501, 384623091229702, 4924836055850737, 21264412149709993, 583141322266592750, 5592755799950501152, 78206057147010955997
Offset: 1
a(2) = 41 because the second and third digits of sqrt(2) are 4 and 1.
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[Floor(Sqrt(2)*10^(n*(n + 1)/2 - 1)) mod (10^n): n in [1..30]]; // Vincenzo Librandi, Feb 15 2016
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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 *)
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a(n) = lift(Mod(floor(sqrt(2)*10^(n*(n + 1)/2 - 1)), 10^n)); \\ G. C. Greubel, Oct 07 2018
Showing 1-5 of 5 results.
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