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 1-5 of 5 results.

A050405 Partial sums of A051879.

Original entry on oeis.org

1, 15, 84, 308, 882, 2142, 4620, 9108, 16731, 29029, 48048, 76440, 117572, 175644, 255816, 364344, 508725, 697851, 942172, 1253868, 1647030, 2137850, 2744820, 3488940, 4393935, 5486481, 6796440, 8357104
Offset: 0

Views

Author

Barry E. Williams, Dec 21 1999

Keywords

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

Crossrefs

Cf. A051879.
Cf. A093644 ((9, 1) Pascal, column m=6).

Programs

  • GAP
    List([0..40], n-> Binomial(n+5, 5)*(3*n+2)/2); # G. C. Greubel, Oct 30 2019
  • Magma
    [Binomial(n+5, 5)*(3*n+2)/2: n in [0..40]]; // G. C. Greubel, Oct 30 2019
    
  • Maple
    seq(binomial(n+5, 5)*(3*n+2)/2, n=0..40); # G. C. Greubel, Oct 30 2019
  • Mathematica
    Accumulate[Accumulate[Table[(n+1)(n+2)(n+3)(9n+4)/24,{n,0,40}]]] (* Harvey P. Dale, Aug 19 2012 *)
  • PARI
    vector(41, n, binomial(n+4, 5)*(3*n-1)/2) \\ G. C. Greubel, Oct 30 2019
    
  • Sage
    [binomial(n+5, 5)*(3*n+2)/2 for n in (0..40)] # G. C. Greubel, Oct 30 2019
    

Formula

a(n) = binomial(n+5, 5)*(3*n + 2)/2.
G.f.: (1+8*x)/(1-x)^7.
E.g.f.: (240 +3360*x +6600*x^2 +4000*x^3 +950*x^4 +92*x^5 +3* x^6) *exp(x)/240. - G. C. Greubel, Oct 30 2019

Extensions

Corrected by T. D. Noe, Nov 09 2006

A005585 5-dimensional pyramidal numbers: a(n) = n*(n+1)*(n+2)*(n+3)*(2n+3)/5!.

Original entry on oeis.org

1, 7, 27, 77, 182, 378, 714, 1254, 2079, 3289, 5005, 7371, 10556, 14756, 20196, 27132, 35853, 46683, 59983, 76153, 95634, 118910, 146510, 179010, 217035, 261261, 312417, 371287, 438712, 515592, 602888, 701624, 812889, 937839, 1077699, 1233765, 1407406
Offset: 1

Views

Author

Keywords

Comments

Convolution of triangular numbers (A000217) and squares (A000290) (n>=1). - Graeme McRae, Jun 07 2006
p^k divides a(p^k-3), a(p^k-2), a(p^k-1) and a(p^k) for prime p > 5 and integer k > 0. p^k divides a((p^k-3)/2) for prime p > 5 and integer k > 0. - Alexander Adamchuk, May 08 2007
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-5) is the number of 6-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
5-dimensional square numbers, fourth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+4, i+4)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Antidiagonal sums of the convolution array A213550. - Clark Kimberling, Jun 17 2012
Binomial transform of (1, 6, 14, 16, 9, 2, 0, 0, 0, ...). - Gary W. Adamson, Jul 28 2015
2*a(n) is number of ways to place 4 queens on an (n+3) X (n+3) chessboard so that they diagonally attack each other exactly 6 times. The maximal possible attack number, p=binomial(k,2)=6 for k=4 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015
While adjusting for offsets, add A000389 to find the next in series A000389, A005585, A051836, A034263, A027800, A051843, A051877, A051878, A051879, A051880, A056118, A271567. (See Bruno Berselli's comments in A271567.) - Bruce J. Nicholson, Jun 21 2018
Coefficients in the terminating series identity 1 - 7*n/(n + 6) + 27*n*(n - 1)/((n + 6)*(n + 7)) - 77*n*(n - 1)*(n - 2)/((n + 6)*(n + 7)*(n + 8)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A040977. - Peter Bala, Feb 18 2019

Examples

			G.f. = x + 7*x^2 + 27*x^3 + 77*x^4 + 182*x^5 + 378*x^6 + 714*x^7 + 1254*x^8 + ... - _Michael Somos_, Jun 24 2018
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = ((-1)^(n+1))*A053120(2*n+3, 5)/16, (1/16 of sixth unsigned column of Chebyshev T-triangle, zeros omitted).
Partial sums of A002415.
Cf. A006542, A040977, A047819, A111125 (third column).
Cf. a(n) = ((-1)^(n+1))*A084960(n+1, 2)/16 (compare with the first line). - Wolfdieter Lang, Aug 04 2014

Programs

  • Magma
    I:=[1, 7, 27, 77, 182, 378]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+6*Self(n-5)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Jun 09 2013
    
  • Maple
    [seq(binomial(n+2,6)-binomial(n,6), n=4..45)]; # Zerinvary Lajos, Jul 21 2006
    A005585:=(1+z)/(z-1)**6; # Simon Plouffe in his 1992 dissertation
  • Mathematica
    With[{c=5!},Table[n(n+1)(n+2)(n+3)(2n+3)/c,{n,40}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{1,7,27,77,182,378},40] (* Harvey P. Dale, Oct 04 2011 *)
    CoefficientList[Series[(1 + x) / (1 - x)^6, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
  • PARI
    a(n)=binomial(n+3,4)*(2*n+3)/5 \\ Charles R Greathouse IV, Jul 28 2015

Formula

G.f.: x*(1+x)/(1-x)^6.
a(n) = 2*C(n+4, 5) - C(n+3, 4). - Paul Barry, Mar 04 2003
a(n) = C(n+3, 5) + C(n+4, 5). - Paul Barry, Mar 17 2003
a(n) = C(n+2, 6) - C(n, 6), n >= 4. - Zerinvary Lajos, Jul 21 2006
a(n) = Sum_{k=1..n} T(k)*T(k+1)/3, where T(n) = n(n+1)/2 is a triangular number. - Alexander Adamchuk, May 08 2007
a(n-1) = (1/4)*Sum_{1 <= x_1, x_2 <= n} |x_1*x_2*det V(x_1,x_2)| = (1/4)*Sum_{1 <= i,j <= n} i*j*|i-j|, where V(x_1,x_2) is the Vandermonde matrix of order 2. First differences of A040977. - Peter Bala, Sep 21 2007
a(n) = C(n+4,4) + 2*C(n+4,5). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), a(1)=1, a(2)=7, a(3)=27, a(4)=77, a(5)=182, a(6)=378. - Harvey P. Dale, Oct 04 2011
a(n) = (1/6)*Sum_{i=1..n+1} (i*Sum_{k=1..i} (i-1)*k). - Wesley Ivan Hurt, Nov 19 2014
E.g.f.: x*(2*x^4 + 35*x^3 + 180*x^2 + 300*x + 120)*exp(x)/120. - Robert Israel, Nov 19 2014
a(n) = A000389(n+3) + A000389(n+4). - Bruce J. Nicholson, Jun 21 2018
a(n) = -a(-3-n) for all n in Z. - Michael Somos, Jun 24 2018
From Amiram Eldar, Jun 28 2020: (Start)
Sum_{n>=1} 1/a(n) = 40*(16*log(2) - 11)/3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 20*(8*Pi - 25)/3. (End)
a(n) = A004302(n+1) - A207361(n+1). - J. M. Bergot, May 20 2022
a(n) = Sum_{i=0..n+1} Sum_{j=i..n+1} i*j*(j-i)/2. - Darío Clavijo, Oct 11 2023
a(n) = (A000538(n+1) - A000330(n+1))/12. - Yasser Arath Chavez Reyes, Feb 21 2024

A093644 (9,1) Pascal triangle.

Original entry on oeis.org

1, 9, 1, 9, 10, 1, 9, 19, 11, 1, 9, 28, 30, 12, 1, 9, 37, 58, 42, 13, 1, 9, 46, 95, 100, 55, 14, 1, 9, 55, 141, 195, 155, 69, 15, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 9, 73, 260, 532, 686, 574, 308, 100, 17, 1, 9, 82, 333, 792, 1218, 1260, 882, 408, 117, 18, 1, 9, 91, 415
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Comments

The array F(9;n,m) gives in the columns m>=1 the figurate numbers based on A017173, including the 11-gonal numbers A051682 (see the W. Lang link).
This is the ninth member, d=9, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5, for d=1..8.
This is an example of a Riordan triangle (see A093560 for a comment and A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group). Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m is G(z,x) = (1+8*z)/(1-(1+x)*z).
The SW-NE diagonals give A022099(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 8. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Triangle T(n,k), read by rows, given by (9,-8,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 10 2011

Examples

			Triangle begins
  [1];
  [9,  1];
  [9, 10,  1];
  [9, 19, 11,  1];
  ...
		

References

  • Kurt Hawlitschek, Johann Faulhaber 1580-1635, Veroeffentlichung der Stadtbibliothek Ulm, Band 18, Ulm, Germany, 1995, Ch. 2.1.4. Figurierte Zahlen.
  • Ivo Schneider: Johannes Faulhaber 1580-1635, Birkhäuser, Basel, Boston, Berlin, 1993, ch.5, pp. 109-122.

Crossrefs

Row sums: A020714(n-1), n >= 1, 1 for n=0, alternating row sums are 1 for n=0, 8 for n=2 and 0 otherwise.
The column sequences give for m=1..9: A017173, A051682 (11-gonal), A007586, A051798, A051879, A050405, A052206, A056117, A056003.
Cf. A093645 (d=10).

Programs

  • Haskell
    a093644 n k = a093644_tabl !! n !! k
    a093644_row n = a093644_tabl !! n
    a093644_tabl = [1] : iterate
                   (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [9, 1]
    -- Reinhard Zumkeller, Aug 31 2014
  • Mathematica
    Join[{1},Table[Binomial[n,k]+8Binomial[n-1,k],{n,20},{k,0,n}]//Flatten] (* Harvey P. Dale, Aug 17 2024 *)

Formula

a(n, m) = F(9;n-m, m) for 0 <= m <= n, otherwise 0, with F(9;0, 0)=1, F(9;n, 0)=9 if n >= 1 and F(9;n, m):=(9*n+m)*binomial(n+m-1, m-1)/m if m >= 1.
Recursion: a(n, m)=0 if m > n, a(0, 0)= 1; a(n, 0)=9 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+8*x)/(1-x)^(m+1), m >= 0.
T(n, k) = C(n, k) + 8*C(n-1, k). - Philippe Deléham, Aug 28 2005
Row n: Expansion of (9+x)*(1+x)^(n-1), n > 0. - Philippe Deléham, Oct 10 2011
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(9 + 19*x + 11*x^2/2! + x^3/3!) = 9 + 28*x + 58*x^2/2! + 100*x^3/3! + 155*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
G.f.: (-1-8*x)/(-1+x+x*y). - R. J. Mathar, Aug 11 2015

A134081 Triangle T(n, k) = binomial(n, k)*((2*k+1)*(n-k) +k+1)/(k+1), read by rows.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 12, 8, 1, 5, 22, 26, 11, 1, 6, 35, 60, 45, 14, 1, 7, 51, 115, 125, 69, 17, 1, 8, 70, 196, 280, 224, 98, 20, 1, 9, 92, 308, 546, 574, 364, 132, 23, 1, 10, 117, 456, 966, 1260, 1050, 552, 171, 26, 1
Offset: 0

Views

Author

Gary W. Adamson, Oct 07 2007

Keywords

Examples

			First few rows of the triangle are:
  1;
  2,  1;
  3,  5,   1;
  4, 12,   8,   1;
  5, 22,  26,  11,  1;
  6, 35,  60,  45, 14,  1;
  7, 51, 115, 125, 69, 17, 1;
  ...
		

Crossrefs

Programs

  • Magma
    A134081:= func< n,k | Binomial(n, k)*((2*k+1)*(n-k) +k+1)/(k+1) >;
    [A134081(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 17 2021
  • Mathematica
    T[n_, k_]:= Binomial[n, k]*((2*k+1)*(n-k) +k+1)/(k+1);
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 17 2021 *)
  • Sage
    def A134081(n,k): return binomial(n, k)*((2*k+1)*(n-k) +k+1)/(k+1)
    flatten([[A134081(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 17 2021
    

Formula

Binomial transform of A112295(unsigned).
From G. C. Greubel, Feb 17 2021: (Start)
T(n, k) = binomial(n, k)*((2*k+1)*(n-k) +k+1)/(k+1).
Sum_{k=0..n} T(n, k) = 2^n *n + 1 = A002064(n). (End)

A271567 Convolution of nonzero triangular numbers (A000217) and nonzero tetradecagonal numbers (A051866).

Original entry on oeis.org

1, 17, 87, 287, 742, 1638, 3234, 5874, 9999, 16159, 25025, 37401, 54236, 76636, 105876, 143412, 190893, 250173, 323323, 412643, 520674, 650210, 804310, 986310, 1199835, 1448811, 1737477, 2070397, 2452472, 2888952, 3385448, 3947944, 4582809, 5296809, 6097119
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 12 2016

Keywords

Comments

More generally, the ordinary generating function for the convolution of triangular numbers and k-gonal numbers is (1 + (k - 3)*x)/(1 - x)^6.

Crossrefs

Cf. similar sequences of the convolution of triangular numbers and k-gonal numbers: A005585 (k=4), A051836 (k=5), A034263 (k=6), A027800 (k=7), A051843 (k=8), A051877 (k=9), A051878 (k=10), A051879 (k=11), A051880 (k=12), A056118 (k=13), this sequence (k=14).

Programs

  • Magma
    /* From definition: */ P:=func; /*, where P(n, k) is the n-th k-gonal number, */ [&+[P(n+1-i, 3)*P(i, 14): i in [1..n]]: n in [1..40]]; // Bruno Berselli, Apr 18 2016
    
  • Magma
    [(n+1)*(n+2)*(n+3)*(n+4)*(12*n+5)/120: n in [0..40]]; // Bruno Berselli, Apr 18 2016
  • Mathematica
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 17, 87, 287, 742, 1638}, 40]
    Table[(n + 1) (n + 2) (n + 3) (n + 4) (12 n + 5)/120, {n, 0, 40}]

Formula

O.g.f.: (1 + 11*x)/(1 - x)^6.
E.g.f.: (120 + 1920*x + 3240*x^2 + 1520*x^3 + 245*x^4 + 12*x^5)*exp(x)/120.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (n + 1)*(n + 2)*(n + 3)*(n + 4)*(12*n + 5)/120.
Sum_{n>=0} 1/a(n) = 20*((15552*(6*log(2) + 3*log(3) + 2*sqrt(3)*log(2 - sqrt(3)) + (2 - sqrt(3))*Pi) - 29449)/531867) = 1.07654258697...

Extensions

Edited by Bruno Berselli, Apr 18 2016
Showing 1-5 of 5 results.