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 1-4 of 4 results.

A223718 Number of unimodal functions [1..n]->[0..2].

Original entry on oeis.org

1, 3, 9, 22, 46, 86, 148, 239, 367, 541, 771, 1068, 1444, 1912, 2486, 3181, 4013, 4999, 6157, 7506, 9066, 10858, 12904, 15227, 17851, 20801, 24103, 27784, 31872, 36396, 41386, 46873, 52889, 59467, 66641, 74446, 82918, 92094, 102012, 112711, 124231
Offset: 0

Views

Author

R. H. Hardin, Mar 26 2013

Keywords

Comments

Column 1 of A223725.

Examples

			Some solutions for n=3
..1....2....0....1....0....2....1....2....0....2....0....1....0....1....0....1
..2....1....1....1....0....0....0....1....0....2....2....1....1....2....2....2
..0....1....0....0....1....0....0....0....2....2....1....1....1....2....0....1
From _Joerg Arndt_, Dec 27 2023: (Start)
The a(3) = 22 such functions are (dots for zeros)
   1:  [ . . . ]
   2:  [ . . 1 ]
   3:  [ . . 2 ]
   4:  [ . 1 . ]
   5:  [ . 1 1 ]
   6:  [ . 1 2 ]
   7:  [ . 2 . ]
   8:  [ . 2 1 ]
   9:  [ . 2 2 ]
  10:  [ 1 . . ]
  11:  [ 1 1 . ]
  12:  [ 1 1 1 ]
  13:  [ 1 1 2 ]
  14:  [ 1 2 . ]
  15:  [ 1 2 1 ]
  16:  [ 1 2 2 ]
  17:  [ 2 . . ]
  18:  [ 2 1 . ]
  19:  [ 2 1 1 ]
  20:  [ 2 2 . ]
  21:  [ 2 2 1 ]
  22:  [ 2 2 2 ]
(End)
		

Crossrefs

Column m=3 of A071920.
Cf. A000124 (unimodal functions [1..n]->[0..1]), A088536 ([1..n] -> [1..n]).

Formula

a(n) = A071920(n,3) = 1+n*(n+1)*(n^2+5*n+18)/24.
G.f.: 1-x*(x^2-3*x+3)*(x^2-x+1) / (x-1)^5 . a(n) = 1+A051744(n). - R. J. Mathar, May 17 2014

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 27 2023

A267370 Partial sums of A140091.

Original entry on oeis.org

0, 6, 21, 48, 90, 150, 231, 336, 468, 630, 825, 1056, 1326, 1638, 1995, 2400, 2856, 3366, 3933, 4560, 5250, 6006, 6831, 7728, 8700, 9750, 10881, 12096, 13398, 14790, 16275, 17856, 19536, 21318, 23205, 25200, 27306, 29526, 31863, 34320, 36900, 39606, 42441, 45408, 48510
Offset: 0

Views

Author

Bruno Berselli, Jan 13 2016

Keywords

Comments

After 0, this sequence is the third column of the array in A185874.
Sequence is related to A051744 by A051744(n) = n*a(n)/3 - Sum_{i=0..n-1} a(i) for n>0.

Examples

			The sequence is also provided by the row sums of the following triangle (see the fourth formula above):
.  0;
.  1,  5;
.  4,  7, 10;
.  9, 11, 13, 15;
. 16, 17, 18, 19, 20;
. 25, 25, 25, 25, 25, 25;
. 36, 35, 34, 33, 32, 31, 30;
. 49, 47, 45, 43, 41, 39, 37, 35;
. 64, 61, 58, 55, 52, 49, 46, 43, 40;
. 81, 77, 73, 69, 65, 61, 57, 53, 49, 45, etc.
First column is A000290.
Second column is A027690.
Third column is included in A189834.
Main diagonal is A008587; other parallel diagonals: A016921, A017029, A017077, A017245, etc.
Diagonal 1, 11, 25, 43, 65, 91, 121, ... is A161532.
		

Crossrefs

Cf. similar sequences of the type n*(n+1)*(n+k)/2: A002411 (k=0), A006002 (k=1), A027480 (k=2), A077414 (k=3, with offset 1), A212343 (k=4, without the initial 0), this sequence (k=5).

Programs

  • Magma
    [n*(n+1)*(n+5)/2: n in [0..50]];
  • Mathematica
    Table[n (n + 1) (n + 5)/2, {n, 0, 50}]
    LinearRecurrence[{4,-6,4,-1},{0,6,21,48},50] (* Harvey P. Dale, Jul 18 2019 *)
  • PARI
    vector(50, n, n--; n*(n+1)*(n+5)/2)
    
  • Sage
    [n*(n+1)*(n+5)/2 for n in (0..50)]
    

Formula

O.g.f.: 3*x*(2 - x)/(1 - x)^4.
E.g.f.: x*(12 + 9*x + x^2)*exp(x)/2.
a(n) = n*(n + 1)*(n + 5)/2.
a(n) = Sum_{i=0..n} n*(n - i) + 5*i, that is: a(n) = A002411(n) + A028895(n). More generally, Sum_{i=0..n} n*(n - i) + k*i = n*(n + 1)*(n + k)/2.
a(n) = 3*A005581(n+1).
a(n+1) - 3*a(n) + 3*a(n-1) = 3*A105163(n) for n>0.
From Amiram Eldar, Jan 06 2021: (Start)
Sum_{n>=1} 1/a(n) = 163/600.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/5 - 253/600. (End)

A202390 Triangle T(n,k), read by rows, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 8, 3, 1, 10, 21, 17, 5, 1, 15, 45, 58, 35, 8, 1, 21, 85, 154, 144, 68, 13, 1, 28, 147, 350, 452, 330, 129, 21, 1, 36, 238, 714, 1195, 1198, 719, 239, 34, 1, 45, 366, 1344, 2799, 3611, 2959, 1506, 436, 55
Offset: 0

Views

Author

Philippe Deléham, Dec 18 2011

Keywords

Comments

T(n,n) = Fibonacci(n+1) = A000045(n+1).
A202390 is jointly generated with A208340 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n-1,x)+x*v(n-1)x and v(n,x)=(x+1)*u(n-1,x)+(x+1)v(n-1,x). The alternating row sums of A202390, and also A208340, are 0 except for the first one. See the Mathematica section. - Clark Kimberling, Feb 27 2012

Examples

			Triangle begins:
  1
  1, 1
  1, 3, 2
  1, 6, 8, 3
  1, 10, 21, 17, 5
  1, 15, 45, 58, 35, 8
  1, 21, 85, 154, 144, 68, 13
  1, 28, 147, 350, 452, 330, 129, 21
		

Crossrefs

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 13;
    u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
    v[n_, x_] := (x + 1)*u[n - 1, x] + (x + 1)*v[n - 1, x];
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]  (* A202390 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]  (* A208340 *)
    Table[u[n, x] /. x -> 1, {n, 1, z}]  (*row sums*)
    Table[u[n, x] /. x -> -1, {n, 1, z}] (*alt. row sums*)

Formula

T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-2) - T(n-2,k) with T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if n
G.f.: (1-x)/(1-(2+y)*x+(1-y^2)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = (-1)^n*A108411(n), A000007(n), A000012(n), A025192(n), A122558(n) for x = -2, -1, 0, 1, 2 respectively.

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

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. A027642, A164555 (Numerators and denominators of Bernoulli numbers).
Cf. A001008, A002805 (Numerators and denominators of harmonic numbers).
Sequences below will be obtained by evaluation of the associated polynomials:

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)
Showing 1-4 of 4 results.