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 1-8 of 8 results.

A032347 Inverse binomial transform of A032346.

Original entry on oeis.org

1, 0, 1, 2, 6, 21, 82, 354, 1671, 8536, 46814, 273907, 1700828, 11158746, 77057021, 558234902, 4230337018, 33448622893, 275322101318, 2354401779494, 20878592918183, 191682453823420, 1819147694792802
Offset: 0

Views

Author

Joe K. Crump (joecr(AT)carolina.rr.com)

Keywords

Links

Crossrefs

Cf. A032346, A046934.

Programs

  • Mathematica
    a[0] = 1; a[1] = 0; a[n_] := a[n] = 1 + Sum[Binomial[n-1, j]*a[j], {j, 2, n-1}]; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Oct 08 2013, after Jon Perry *)
nmax = 20; Assuming[x > 0, CoefficientList[Series[E^(E^x) * (1/E + ExpIntegralEi[-1] - ExpIntegralEi[-E^x]), {x, 0, nmax}], x] ] * Range[0, nmax]! (* Vaclav Kotesovec, Jul 10 2020 *)
  • PARI
    {a(n)=local(A=1+x*O(x^n)); for(i=0, n, A=1 - x * (1 - subst(A, x, x/(1-x)) / (1 - x))); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", ")) \\ Vaclav Kotesovec, Jul 10 2020

Formula

E.g.f. satisfies A' = exp(x) A - 1.
Recurrence: a(1)=0, a(2)=1, for n > 2, a(n) = 1 + Sum_{j=2..n-1} binomial(n-1, j)*a(j). - Jon Perry, Apr 26 2005
G.f. A(x) satisfies: A(x) = 1 - x * (1 - A(x/(1 - x)) / (1 - x)). - Ilya Gutkovskiy, Jul 10 2020

A046934 Same rule as Aitken triangle (A011971) except a(0,0)=1, a(1,0)=0.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 2, 3, 4, 6, 6, 8, 11, 15, 21, 21, 27, 35, 46, 61, 82, 82, 103, 130, 165, 211, 272, 354, 354, 436, 539, 669, 834, 1045, 1317, 1671, 1671, 2025, 2461, 3000, 3669, 4503, 5548, 6865, 8536, 8536, 10207, 12232, 14693, 17693, 21362, 25865, 31413, 38278, 46814
Offset: 0

Views

Author

N. J. A. Sloane, R. K. Guy

Keywords

Comments

First differences of A046935 = this triangle seen as flattened list without the initial term. - Reinhard Zumkeller, Nov 10 2013

Examples

			[0] [   1]
[1] [   0,     1]
[2] [   1,     1,     2]
[3] [   2,     3,     4,     6]
[4] [   6,     8,    11,    15,    21]
[5] [  21,    27,    35,    46,    61,    82]
[6] [  82,   103,   130,   165,   211,   272,   354]
[7] [ 354,   436,   539,   669,   834,  1045,  1317,  1671]
[8] [1671,  2025,  2461,  3000,  3669,  4503,  5548,  6865,  8536]
[9] [8536, 10207, 12232, 14693, 17693, 21362, 25865, 31413, 38278, 46814]

Links

Crossrefs

Borders give A032347 and A032346. Cf. A046935.

Programs

  • Haskell
    a046934 n k = a046934_tabl !! n !! k
a046934_row n = a046934_tabl !! n
a046934_tabl = [1] : iterate (\row -> scanl (+) (last row) row) [0,1]
a046934_list = concat a046934_tabl
-- Reinhard Zumkeller, Nov 10 2013
  • Maple
    alias(PS = ListTools:-PartialSums):
A046934Triangle := proc(len) local a, k, P, T; a := 0; P := [1]; T := [];
for k from 1 to len  do T := [op(T), P]; P := PS([a, op(P)]); a := P[-1] od;
ListTools:-Flatten(T) end: A046934Triangle(10);  # Peter Luschny, Mar 29 2022
  • Mathematica
    a[0, 0] = 1; a[1, 0] = 0; a[n_, 0] := a[n, 0] = a[n-1, n-1]; a[n_, k_] := a[n, k] = a[n, k-1] + a[n-1, k-1]; Table[a[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 14 2013 *)

A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .

Original entry on oeis.org

1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11
Offset: 0

Views

Author

Thomas Scheuerle, Feb 13 2022

Keywords

Comments

If we want to calculate the sum of finite differences for a sequence b(n):
b(0)*T(0, n) + ... + b(n)*T(n, n) = b(0) + b(1) + ... + b(n) + (b(1) - b(0)) + ... + (b(n) - b(n-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... This sum includes the sequence b(n) itself. This defines an invertible linear sequence transformation with a deep connection to Bernoulli numbers and other interesting sequences of rational numbers.
From Thomas Scheuerle, Apr 29 2024: (Start)
These are the coefficients of the polynomials defined by the recurrence: P(k, x) = P(k - 1, x) + (x^2 - x)*P(k - 2, x) + 1, with P(-1, x) = 0 and P(0, x) = 1. This can also be expressed as P(k, x) = Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^2 - x)^(m - 1) = Sum_{n=0..k} T(n, k)*x^(k-n). If we would evaluate P(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1+(t-1)*x)*(1 - t*x)).
We may replace (x^2 - x) by (x^(-2) - x^(-1)) to get the coefficients in reverse order: x^k*Sum_{m=1..k+1} binomial(k+2 - m, m)*(x^(-2) - x^(-1))^(m - 1) = Sum_{n=0..k} T(n, k)*x^n = F(k, x). If we would evaluate F(k, t) as sequence for some fixed t then we get the expansion of 1/((1 - x)*(1 - (t-1)*x)*(1 - t*x)). (End)

Examples

			Triangle begins with T(n, k):
   n=   0,  1,   2,   3,   4,   5,   6,   7,   8
  k=0   1
  k=1   0,  2
  k=2   1, -1,   3
  k=3   0,  3,  -3,   4
  k=4   1, -2,   7,  -6,   5
  k=5   0,  4,  -8,  14, -10,   6
  k=6   1, -3,  13, -21,  25, -15,   7
  k=7   0,  5, -15,  35, -45,  41, -21,   8
  k=8   1, -4,  21, -49,  81, -85,  63, -28,   9
  ...

Crossrefs

Cf. A000027, A000032, A000045, A000110, A000217.
Cf. A001045, A004006, A032346, A051744, A052875.
Cf. A084416, A130595, A145766.
Cf. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Cf. A344037, A372245.
Sequences below will be obtained by evaluation of the associated polynomials:
Cf. A000295, A000392, A000975, A016208.
Cf. A016218, A016228, A016241, A016249.
Cf. A016256, A016261, A249997.

Programs

  • PARI
    A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))
  • PARI
    A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024

Formula

b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
= Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
= Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
Sum_{n=0..k} T(n, k)*A032346(n) = A032346(k+1).
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*A000110(n+1) = A000110(k+2) - 1.
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
Sum_{n=0..k} T(n, k)*A000032(n) = A000032(2*floor(k/2)+2) - 2. A000032 are the Lucas numbers.
Sum_{n=0..k} T(n, k)*A001045(n) = A145766(floor((k+1)/2)). A001045 is the Jacobsthal sequence.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)

A275872 A binomial convolution recurrence sequence.

Original entry on oeis.org

0, 0, 1, 1, 2, 6, 18, 54, 173, 605, 2274, 9020, 37486, 163128, 743101, 3535765, 17518018, 90126158, 480514430, 2650912738, 15112253425, 88903779401, 539003066674, 3363608949132, 21581457167994, 142227480847092, 961868098767105, 6669657795455817, 47380035801732034, 344555811578909254, 2563218995058696890
Offset: 0

Views

Author

Olivier Gérard, Aug 11 2016

Keywords

Comments

Shifts 2 places left and decreases by one under a variant of binomial transform (see formula section).

Links

Crossrefs

Cf. A000994, A007476, A032346.

Programs

  • Maple
    A[0]:= 0:
A[1]:= 0:
for m from 2 to 50 do
  A[m]:= 1 + add(binomial(m-1,i+1)*A[i],i=0..m-2)
od:
seq(A[i],i=0..50); # Robert Israel, Aug 28 2016
  • Mathematica
    Clear[a]; a[0] = 0 ; a[1] = 0; a[n_] := a[n] = 1 + Sum[Binomial[n - 1, j+1]*a[j], {j, 0, n - 1}]; Table[a[n], {n, 0, 22}]
  • PARI
    first(n)=my(v=vector(n)); for(k=0,n-2, v[k+2]=sum(i=2,k, binomial(k+1,i+1)*v[i])+1); concat(0,v) \\ Charles R Greathouse IV, Aug 29 2016

Formula

Sum_{i=0..n} binomial(n+1,i+1)*a(i) = a(n+2) - 1.
G.f. g(x) satisfies g(x) = x^2/(1-x) + x^2*g(x/(1-x))/(1-x)^2. - Robert Israel, Aug 28 2016

A309495 Triangle read by rows, derived from A007318, row sums = the Bell Sequence.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 5, 6, 1, 4, 9, 17, 21, 1, 5, 14, 34, 67, 82, 1, 6, 20, 58, 148, 290, 354, 1, 7, 27, 90, 275, 701, 1368, 1671, 1, 8, 35, 131, 460, 1411, 3579, 6986, 8536, 1, 9, 44, 182, 716, 2536, 7738, 19620, 38315, 46814, 1, 10, 54, 244, 1057, 4213, 14846, 45251, 114798, 224189, 273907
Offset: 1

Views

Author

Gary W. Adamson, Aug 04 2019

Keywords

Comments

As described in A160185, we extract eigensequences of a rotated variant of Pascal's triangle:
1;
3, 1;
3, 2, 1;
1, 1, 1, 1;
Say, for these 4 columns, the eigensequence is (1, 4, 9, 15). Then preface the latter with a zero and take the first finite difference row, = (1, 3, 5, 6), fourth row of the triangle.

Examples

			Row 5 of A121207 is (1, 5, 14, 31, 52). Preface with a zero and take the first difference row:
     (0,  1,  5, 14, 31, 52)
  (..., 1,  4,  9, 17, 21) = row 5 of the triangle.
First few rows of the triangle:
  1;
  1, 1;
  1, 2,  2;
  1, 3,  5,  6;
  1, 4,  9, 17,  21;
  1, 5, 14, 34,  67,  82;
  1, 6, 20, 58, 148, 290, 354;
  ...

Links

Crossrefs

Row sums are A000110.
Main diagonal is A032346.
Cf. A007318, A121207, A160185.

Programs

  • PARI
    \\ here U(n) is A121207.
U(n)={my(M=matrix(n,n)); for(n=1, n, M[n,1]=1; for(k=1, n-1, M[n,k+1]=sum(j=1, k, M[n-j, k-j+1]*binomial(n-2,j-1)))); M}
T(n)={my(A=U(n+1)); vector(n, n, my(t=A[n+1,2..n+1]); t-concat([0], t[1..n-1]))}
{ my(A=T(10)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Feb 20 2022

Formula

T(n,k) = A121207(n,k) - A121207(n, k-1) for k >= 2.

Extensions

Terms a(37) and beyond from Andrew Howroyd, Feb 20 2022

A352861 a(n) = 1 + Sum_{k=0..n-1} binomial(n+2,k+3) * a(k).

Original entry on oeis.org

1, 2, 7, 28, 121, 570, 2911, 15968, 93433, 580162, 3806275, 26284368, 190415809, 1442982350, 11409436363, 93913277608, 803094241309, 7121757279798, 65383520552131, 620517308328812, 6079168380979213, 61402851498255790, 638674759049919079, 6833589979500278700
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 06 2022

Keywords

Crossrefs

Cf. A000110, A032346, A032347, A045499, A045501, A186021, A352862.

Programs

  • Mathematica
    a[n_] := a[n] = 1 + Sum[Binomial[n + 2, k + 3] a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 23}]
nmax = 23; A[] = 0; Do[A[x] = 1/(1 - x) + x A[x/(1 - x)]/(1 - x)^4 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * A(x/(1 - x)) / (1 - x)^4.
a(0) = 1; a(n) = a(n-1) + Sum_{k=0..n-1} binomial(n+1,k+2) * a(k).

A352862 a(n) = 1 + Sum_{k=0..n-1} binomial(n+3,k+4) * a(k).

Original entry on oeis.org

1, 2, 8, 36, 170, 865, 4742, 27757, 172375, 1130865, 7809057, 56572404, 428710587, 3389749264, 27901667938, 238599540142, 2115876327408, 19425465343555, 184355895494512, 1806122902809371, 18242807108024625, 189750478368293523, 2030261803964224359, 22323607721661782198
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 06 2022

Keywords

Crossrefs

Cf. A000110, A032346, A032347, A045499, A045500, A186021, A352861.

Programs

  • Mathematica
    a[n_] := a[n] = 1 + Sum[Binomial[n + 3, k + 4] a[k], {k, 0, n - 1}]; Table[a[n], {n, 0, 23}]
nmax = 23; A[] = 0; Do[A[x] = 1/(1 - x) + x A[x/(1 - x)]/(1 - x)^5 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

Formula

G.f. A(x) satisfies: A(x) = 1 / (1 - x) + x * A(x/(1 - x)) / (1 - x)^5.
a(0) = 1; a(n) = a(n-1) + Sum_{k=0..n-1} binomial(n+2,k+3) * a(k).

A275871 Row sums and second diagonal of A046934.

Original entry on oeis.org

1, 1, 4, 15, 61, 272, 1317, 6865, 38278, 227093, 1426921, 9457918, 65898275, 481177881, 3672102116, 29218285875, 241873478425, 2079079678176, 18524191138689, 170803860905237, 1627465240969382
Offset: 0

Views

Author

Olivier Gérard, Aug 11 2016

Keywords

Comments

The offset corresponds to the definition of A046934.
Differences of A032346 and of A032347.

Crossrefs

Cf. A046934, A046935, A032346, A032347.
Cf. A005493 (first differences of Bell numbers, second diagonal and row sum of A011971).

Programs

  • Mathematica
    Clear[d]; d[0] = 1; d[1] = 0; d[n_] := d[n] =
  1 + Sum[Binomial[n - 1, j]*d[j], {j, 2, n - 1}]; Table[
d[n + 2] - d[n + 1], {n, 0, 22}] (* From the code by  J.-F. Alcover and Jon Perry in A032347 *)
Showing 1-8 of 8 results.

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