cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A003261 Woodall (or Riesel) numbers: n*2^n - 1.

Original entry on oeis.org

1, 7, 23, 63, 159, 383, 895, 2047, 4607, 10239, 22527, 49151, 106495, 229375, 491519, 1048575, 2228223, 4718591, 9961471, 20971519, 44040191, 92274687, 192937983, 402653183, 838860799, 1744830463, 3623878655, 7516192767, 15569256447, 32212254719, 66571993087
Offset: 1

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Author

Keywords

Comments

For n>1, a(n) is base at which zero is reached for the function "write f(j) in base j, read as base j+1 and then subtract 1 to give f(j+1)" starting from f(n) = n^2 - 1. - Henry Bottomley, Aug 06 2000
Sequence corresponds also to the maximum chain length of the classic puzzle whereby, under agreed commercial terms, an asset of unringed golden chain, when judiciously fragmented into as few as n pieces and n-1 opened links (through n-1 cuts), might be used to settle debt sequentially, with a golden link covering for unit cost. Here beside the n-1 opened links, the n fragmented pieces have lengths n, 2*n, 4*n, ..., 2^(n-1)*n. For instance, the chain of original length a(5)=159, if segregated by 4 cuts into 5+1+10+1+20+1+40+1+80, may be used to pay sequentially, i.e., a link-cost at a time, for an equivalent cost up to 159 links, to the same creditor. - Lekraj Beedassy, Feb 06 2003

Examples

			G.f. = x + 7*x^2 + 23*x^3 + 63*x^4 + 159*x^5 + 383*x^6 + 895*x^7 + ... - _Michael Somos_, Nov 04 2018
		

References

  • A. Brousseau, Number Theory Tables. Fibonacci Association, San Jose, CA, 1973, p. 159.
  • K. R. Bhutani and A. B. Levin, "The Problem of Sawing a Chain", Journal of Recreational Mathematics 2002-3 31(1) 32-35.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • M. Gardner, Martin Gardner's Sixth Book of Mathematical Diversions from Scientific American, "Gold Links", Problem 4, pp. 50-51; 57-58, University of Chicago Press, 1983.
  • O. O'Shea, Mathematical Brainteasers with Surprising Solutions, Problem 76, pp. 183-185, Prometheus Books, Guilford, Connecticut, 2020.
  • Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 241.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A036289(n) - 1 = A002064(n) - 2.
Cf. A133653.

Programs

Formula

G.f.: x*(-1-2*x+4*x^2) / ( (x-1)*(-1+2*x)^2 ). - Simon Plouffe in his 1992 dissertation
Binomial transform of A133653 and double binomial transform of [1, 5, -1, 1, -1, 1, ...]. - Gary W. Adamson, Sep 19 2007
a(n) = -(2)^n * A006127(-n) for all n in Z. - Michael Somos, Nov 04 2018
E.g.f.: 1 + exp(x)*(2*exp(x)*x - 1). - Stefano Spezia, Nov 24 2024

A175553 Product of first k triangular numbers divided by the sum of first k triangular numbers is an integer.

Original entry on oeis.org

1, 4, 7, 8, 10, 12, 13, 14, 16, 18, 19, 20, 22, 23, 24, 25, 26, 28, 30, 31, 32, 33, 34, 36, 37, 38, 40, 42, 43, 44, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 58, 60, 61, 62, 63, 64, 66, 67, 68, 70, 72, 73, 74, 75, 76, 78, 79, 80, 82, 83, 84, 85, 86, 88, 89, 90, 91, 92, 93, 94, 96
Offset: 1

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Author

Ctibor O. Zizka, Jun 26 2010

Keywords

Comments

Numbers k such that (1*3*6*10* ... *(k*(k+1)/2)) / (1+3+6+10+ ... +(k*(k+1)/2)) is an integer. What if, instead of triangular numbers, we use squares, 1*4*...*(k*k) / (1+4+...+k*k); odd numbers, 1*3*...*(2*k-1) / (1+3+...+(2*k-1)); or Fibonacci numbers, F(1)* ... *F(k) / (F(1)+ ... + F(k))?
It appears that the corresponding sequence for the Fibonacci numbers is given in A133653. - John W. Layman, Jul 10 2010
k > 6 is in this sequence if and only if k+2 is composite. - Robert Israel, Nov 04 2021

Examples

			For k=4 we have 1*3*6*10 /(1+3+6+10) = 9 so k=4 belongs to the sequence.
		

Crossrefs

Cf. A133653. - John W. Layman, Jul 10 2010

Programs

  • Maple
    A006472 := proc(n) n!*(n-1)!/2^(n-1) ; end proc:
    A000292 := proc(n) binomial(n+2,3) ; end proc:
    for n from 1 to 200 do a := A006472(n+1)/A000292(n) ; if type(a,'integer') then printf("%d,",n) ; end if; end do: # R. J. Mathar, Jun 28 2010
  • Mathematica
    fQ[n_] := Mod[6n!(n - 1)!, (n + 2)2^n ] == 0; Select[Range@ 96, fQ@# &] (* Robert G. Wilson v, Jun 29 2010 *)

Formula

{k: A006472(k+1)/A000292(k) in Z}. - R. J. Mathar, Jun 28 2010

Extensions

More terms from R. J. Mathar and Robert G. Wilson v, Jun 28 2010

A239977 a(n) = -Sum_{k=0..n} binomial(n, k)*A226158(k).

Original entry on oeis.org

0, 1, 3, 6, 9, 10, 9, 14, 33, 18, -135, 22, 2097, 26, -38199, 30, 929601, 34, -28820583, 38, 1109652945, 42, -51943281687, 46, 2905151042529, 50, -191329672483911, 54, 14655626154768753, 58, -1291885088448017655, 62, 129848163681107302017, 66
Offset: 0

Views

Author

Paul Curtz, Mar 30 2014

Keywords

Comments

Let T(n, k) denote the difference table of a(n).
(-1)^(k+1)*T(3, 3+k) = T(k+3, 3) for k >= 0.
Without the first two rows and the first two columns we have the core of the Genocchi numbers, like A240581(n)/A239315(n) for the Bernoulli numbers. See A226158(n).
0, 1, 3, 6, 9, 10, ...
1, 2, 3, 3, 1, -1, ...
1, 1, 0, -2, -2, 6, ...
0, -1, -2, 0, 8, 8, ...
-1, -1, 2, 8, 0, -56, ...
0, 3, 6, -8, -56, 0, ...
The definition reflects the identity 2*((1-2^n)*B(n,1) + n) =
2*Sum_{k=0..n} C(n,k)*(2^k-1)*B(k,1) where B(n,x) denotes the Bernoulli polynomials. - Peter Luschny, Apr 16 2014

Crossrefs

Cf. A083007.

Programs

  • Magma
    [0,1] cat [2*((1 - 2^n)*Bernoulli(n) + n): n in [2..40]]; // Vincenzo Librandi, Mar 03 2015
  • Maple
    A239977 := n -> 2*((1-2^n)*bernoulli(n,1) + n):
    seq(A239977(n), n=0..33); # Peter Luschny, Mar 08 2015
  • Mathematica
    a[n_] := (EulerE[n-1, 0]+2)*n; a[0] = 0; a[1] = 1; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Apr 02 2014 *)

Formula

a(n) = 2*n + A226158(n).
a(2n) is divisible by 3.
a(2n+1) = A133653(n).
a(n) = 2*((1 - 2^n)*B(n, 1) + n), B(n, x) the Bernoulli polynomial. - Peter Luschny, Apr 16 2014
From Peter Bala, Mar 02 2015: (Start)
a(n) = (-2)^n * ( B(n,-1/2) - B(n,0) ), where B(n,x) denotes the n-th Bernoulli polynomial. More generally, for any nonzero integer k, k^n*( B(n,1/k) - B(n,0) ) is an integer for n >= 0. Cf. A083007.
a(0) = 0 and for n >= 1, a(n) = 1 - 1/(n + 1)*Sum_{k = 1..n-1} (-2)^(n-k)*binomial(n+1,k)*a(k).
E.g.f.: 2*x*exp(2*x)/(1 + exp(x)) = x + 3*x^2/2! + 6*x^3/3! + .... (End)
a(n) = Sum_{k=0..n-1} 2^k*binomial(n, k)*Bernoulli(k, 1). - Peter Luschny, Aug 17 2021
Showing 1-3 of 3 results.