A093564 -id:A093564 - cp's OEIS frontend

A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n(7n-5)/2.

0, 1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 396, 474, 559, 651, 750, 856, 969, 1089, 1216, 1350, 1491, 1639, 1794, 1956, 2125, 2301, 2484, 2674, 2871, 3075, 3286, 3504, 3729, 3961, 4200, 4446, 4699, 4959, 5226, 5500, 5781, 6069, 6364

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 9, ... and the parallel line from 1, in the direction 1, 24, ..., in the square spiral whose vertices are the generalized 9-gonal (enneagonal) numbers A118277. Also sequence found by reading the same lines in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 10 2011
Number of ordered pairs of integers (x,y) with abs(x) < n, abs(y) < n and x+y <= n. - Reinhard Zumkeller, Jan 23 2012
Partial sums give A007584. - Omar E. Pol, Jan 15 2013

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe and William A. Tedeschi, Table of n, a(n) for n = 0..10000 (1000 terms were computed by T. D. Noe)
S. Barbero, U. Cerruti and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7, section 4.4.
C. K. Cook and M. R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 343
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Nonagonal Number.
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A093564 ((7, 1) Pascal, column m=2). Partial sums of A016993.

Cf. A131875, A057655, A069099, A244646.

a001106 n = length [(x,y) | x <- [-n+1..n-1], y <- [-n+1..n-1], x + y <= n]
-- Reinhard Zumkeller, Jan 23 2012

a001106 n = n*(7*n-5) `div` 2 -- James Spahlinger, Oct 18 2012

Table[n(7n - 5)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 9}, 50] (* Harvey P. Dale, Nov 06 2011 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[9], n], {n, 0, 43}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
PolygonalNumber[9,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 19 2019 *)

a(n)=n*(7*n-5)/2 \\ Charles R Greathouse IV, Jun 10 2011

def aList(): # Intended to compute the initial segment of the sequence, not isolated terms.
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 7, y + 7
A001106 = aList()
print([next(A001106) for i in range(49)]) # Peter Luschny, Aug 04 2019

a(n) = (7*n - 5)*n/2.
G.f.: x*(1+6*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation.
a(n) = n + 7*A000217(n-1). - Floor van Lamoen, Oct 14 2005
Starting (1, 9, 24, 46, 75, ...) gives the binomial transform of (1, 8, 7, 0, 0, 0, ...). - Gary W. Adamson, Jul 22 2007
Row sums of triangle A131875 starting (1, 9, 24, 46, 75, 111, ...). A001106 = binomial transform of (1, 8, 7, 0, 0, 0, ...). - Gary W. Adamson, Jul 22 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0) = 0, a(1) = 1, a(2) = 9. - Jaume Oliver Lafont, Dec 02 2008
a(n) = 2*a(n-1) - a(n-2) + 7. - Mohamed Bouhamida, May 05 2010
a(n) = a(n-1) + 7*n - 6 (with a(0) = 0). - Vincenzo Librandi, Nov 12 2010
a(n) = A174738(7n). - Philippe Deléham, Mar 26 2013
a(7*a(n) + 22*n + 1) =  a(7*a(n) + 22*n) + a(7*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: x*(2 + 7*x)*exp(x)/2. - Ilya Gutkovskiy, Jul 28 2016
a(n+2) + A000217(n) = (2*n+3)^2. - Ezhilarasu Velayutham, Mar 18 2020
Product_{n>=2} (1 - 1/a(n)) = 7/9. - Amiram Eldar, Jan 21 2021
Sum_{n>=1} 1/a(n) = A244646. - Amiram Eldar, Nov 12 2021
a(n) = A000217(3*n-2) - (n-1)^2. - Charlie Marion, Feb 27 2022
a(n) = 3*A000217(n) + 2*A005563(n-2). In general, if P(k,n) = the n-th k-gonal number, then P(m*k,n) = m*P(k,n) + (m-1)*A005563(n-2). - Charlie Marion, Feb 21 2023

A029653 Numbers in (2,1)-Pascal triangle (by row).

Original entry on oeis.org

1, 2, 1, 2, 3, 1, 2, 5, 4, 1, 2, 7, 9, 5, 1, 2, 9, 16, 14, 6, 1, 2, 11, 25, 30, 20, 7, 1, 2, 13, 36, 55, 50, 27, 8, 1, 2, 15, 49, 91, 105, 77, 35, 9, 1, 2, 17, 64, 140, 196, 182, 112, 44, 10, 1, 2, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 2, 21, 100, 285

Offset: 0

Mohammad K. Azarian

Reverse of A029635. Row sums are A003945. Diagonal sums are Fibonacci(n+2) = Sum_{k=0..floor(n/2)} (2n-3k)*C(n-k,n-2k)/(n-k). - Paul Barry, Jan 30 2005
Riordan array ((1+x)/(1-x), x/(1-x)). The signed triangle (-1)^(n-k)T(n,k) or ((1-x)/(1+x), x/(1+x)) is the inverse of A055248. Row sums are A003945. Diagonal sums are F(n+2). - Paul Barry, Feb 03 2005
Row sums = A003945: (1, 3, 6, 12, 24, 48, 96, ...) = (1, 3, 7, 15, 31, 63, 127, ...) - (0, 0, 1, 3, 7, 15, 31, ...); where (1, 3, 7, 15, ...) = A000225. - Gary W. Adamson, Apr 22 2007
Triangle T(n,k), read by rows, given by (2,-1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 17 2011
A029653 is jointly generated with A208510 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)=u(n-1,x)+x*v(n-1,x)+1. See the Mathematica section. - Clark Kimberling, Feb 28 2012
For a closed-form formula for arbitrary left and right borders of Pascal like triangle, see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle, see A228576. - Boris Putievskiy, Sep 04 2013
The n-th row polynomial is (2 + x)*(1 + x)^(n-1) for n >= 1. More generally, the n-th row polynomial of the Riordan array ( (1-a*x)/(1-b*x), x/(1-b*x) ) is (b - a + x)*(b + x)^(n-1) for n >= 1. - Peter Bala, Feb 25 2018

			The triangle T(n,k) begins:
n\k 0  1  2   3   4   5   6   7  8  9 10 ...
0:  1
1:  2  1
2:  2  3  1
3:  2  5  4   1
4:  2  7  9   5   1
5:  2  9 16  14   6   1
6:  2 11 25  30  20   7   1
7:  2 13 36  55  50  27   8   1
8:  2 15 49  91 105  77  35   9  1
9:  2 17 64 140 196 182 112  44 10  1
10: 2 19 81 204 336 378 294 156 54 11  1
... Reformatted. - _Wolfdieter Lang_, Jan 09 2015
With the array M(k) as defined in the Formula section, the infinite product M(0)*M(1)*M(2)*... begins
/1        \/1         \/1        \      /1        \
|2 1      ||0 1       ||0 1      |      |2 1      |
|2 1 1    ||0 2 1     ||0 0 1    |... = |2 3 1    |
|2 1 1 1  ||0 2 1 1   ||0 0 2 1  |      |2 5 4 1  |
|2 1 1 1 1||0 2 1 1 1 ||0 0 2 1 1|      |2 7 9 5 1|
|...      ||...       ||...      |      |...      |
- _Peter Bala_, Dec 27 2014

Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Mohammad K. Azarian, Identities Involving Lucas or Fibonacci and Lucas Numbers as Binomial Sums, International Journal of Contemporary Mathematical Sciences, Vol. 7, No. 45, 2012, pp. 2221-2227.
Paul Barry, A Note on a Family of Generalized Pascal Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.4.
Hacene Belbachir and Athmane Benmezai, Expansion of Fibonacci and Lucas Polynomials: An Answer to Prodinger's Question, Journal of Integer Sequences, Vol. 15 (2012), #12.7.6.
Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids, English translation published by Fibonacci Association, Santa Clara Univ., Santa Clara, CA, 1993; see p. 39.
H. Hosoya, Pascal's triangle, non-adjacent numbers and D-dimensional atomic orbitals, J. Math. Chemistry, vol. 23, 1998, 169-178.
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 13 2013
M. Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Huyile Liang, Yanni Pei, and Yi Wang, Analytic combinatorics of coordination numbers of cubic lattices, arXiv:2302.11856 [math.CO], 2023. See p. 8.
Mark C. Wilson, Asymptotics for generalized Riordan arrays. International Conference on Analysis of Algorithms DMTCS proc. AD. Vol. 323. 2005.

(d, 1) Pascal triangles: A007318(d=1), A093560(3), A093561(4), A093562(5), A093563(6), A093564(7), A093565(8), A093644(9), A093645(10).
Cf. A003945, A208510, A228196, A228576.
Cf. A078812, A106195.

a029653 n k = a029653_tabl !! n !! k
a029653_row n = a029653_tabl !! n
a029653_tabl = [1] : iterate
               (\xs -> zipWith (+) ([0] ++ xs) (xs ++ [0])) [2, 1]
-- Reinhard Zumkeller, Dec 16 2013

A029653 :=  proc(n,k)
if n = 0 then
  1;
else
  binomial(n-1, k)+binomial(n, k)
fi
end proc: # R. J. Mathar, Jun 30 2013

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + x*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
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[%]  (* A208510 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A029653 *)
(* Clark Kimberling, Feb 28 2012 *)

from sympy import Poly
from sympy.abc import x
def u(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + x*v(n - 1, x) + 1
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 27 2017

from math import comb, isqrt
def A029653(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*((r<<1)-a)//r if n else 1 # Chai Wah Wu, Nov 12 2024

T(n, k) = C(n-2, k-1) + C(n-2, k) + C(n-1, k-1) + C(n-1, k) except for n=0.
G.f.: (1 + x + y + xy)/(1 - y - xy). - Ralf Stephan, May 17 2004
T(n, k) = (2n-k)*binomial(n, n-k)/n, n, k > 0. - Paul Barry, Jan 30 2005
Sum_{k=0..n} T(n, k)*x^k gives A003945-A003954 for x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, Jul 10 2005
T(n, k) = C(n-1, k) + C(n, k). - Philippe Deléham, Jul 10 2005
Equals A097806 * A007318, i.e., the pairwise operator * Pascal's Triangle as infinite lower triangular matrices. - Gary W. Adamson, Apr 22 2007
From Peter Bala, Dec 27 2014: (Start)
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 5*x + 4*x^2/2! + x^3/3!) = 2 + 7*x + 16*x^2/2! + 30*x^3/3! + 50*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ).
Let M denote the lower unit triangular array with 1's on the main diagonal and 1's everywhere else below the main diagonal except for the first column which consists of the sequence [1,2,2,2,...]. For k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/ having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite product M(0)*M(1)*M(2)*... (which is clearly well-defined). See the Example section. (End)

More terms from James Sellers

A016993 a(n) = 7*n + 1.

Original entry on oeis.org

1, 8, 15, 22, 29, 36, 43, 50, 57, 64, 71, 78, 85, 92, 99, 106, 113, 120, 127, 134, 141, 148, 155, 162, 169, 176, 183, 190, 197, 204, 211, 218, 225, 232, 239, 246, 253, 260, 267, 274, 281, 288, 295, 302, 309, 316, 323, 330, 337, 344, 351, 358, 365, 372, 379

Offset: 0

N. J. A. Sloane

For n > 3, also the number of (not necessarily maximal) cliques in the n-web graph. - Eric W. Weisstein, Nov 29 2017

The number of notes in a musical scale of n octaves. - Geoffrey Trueman Falk, Feb 16 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Clique
Eric Weisstein's World of Mathematics, Web Graph
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A093564 (column 1).

Cf. A131844, A008589.

a016993 = (+ 1) . (* 7)
a016993_list = [1, 8 ..]  -- Reinhard Zumkeller, Jan 25 2013

[7*n+1: n in [0..60]]; // Vincenzo Librandi, May 28 2011

A016993:=n->7*n+1: seq(A016993(n), n=0..70); # Wesley Ivan Hurt, Nov 01 2014

7*Range[0, 55] + 1 (* Alonso del Arte, Oct 26 2014 *)
Table[7 n + 1, {n, 0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
LinearRecurrence[{2, -1}, {8, 15}, {0, 20}] (* Eric W. Weisstein, Nov 29 2017 *)
CoefficientList[Series[(1 + 6 x)/(-1 + x)^2, {x, 0, 20}], x] (* Eric W. Weisstein, Nov 29 2017 *)

a(n)=7*n+1 \\ Charles R Greathouse IV, Jul 10 2016

a(n) = 7*n + 1.
G.f.: (1+6*x)/(1-x)^2.
From Elmo R. Oliveira, Mar 07 2024: (Start)
a(n) = 2*a(n-1) - a(n-2).
E.g.f.: (1 + 7*x)*exp(x). (End)

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A022097 Fibonacci sequence beginning 1, 7.

Original entry on oeis.org

1, 7, 8, 15, 23, 38, 61, 99, 160, 259, 419, 678, 1097, 1775, 2872, 4647, 7519, 12166, 19685, 31851, 51536, 83387, 134923, 218310, 353233, 571543, 924776, 1496319, 2421095, 3917414, 6338509, 10255923, 16594432, 26850355, 43444787, 70295142, 113739929

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(7;n-1-k,k) with n>=1, a(-1)=6. These are the SW-NE diagonals in P(7;n,k), the (7,1) Pascal triangle A093564. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
Pisano period lengths: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 18, 60, ... (perhaps the same as A001175). - R. J. Mathar, Aug 10 2012
For n >= 1, a(n) is the number of edge covers of the tadpole graph T_{4,n-1} with T_{4,0} interpreted as just the cycle graph C_4. Example: If n=2, we have C_4 and path P_1 joined by a bridge. This is the cycle with a pendant and has 7 edge covers. - Feryal Alayont, Sep 22 2024

G. C. Greubel, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Tadpole Graph.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000045, A001175, A093564, A101220, A109754, A118654, A131778.

a0:=1; a1:=7; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

First /@ NestList[{Last@ #, Total@ #} &, {1, 7}, 36] (* or *)
CoefficientList[Series[(1 + 6 x)/(1 - x - x^2), {x, 0, 36}], x] (* Michael De Vlieger, Feb 20 2017 *)
LinearRecurrence[{1,1},{1,7},40] (* Harvey P. Dale, May 17 2018 *)

a(n)=([0,1; 1,1]^n*[1;7])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

A022097=BinaryRecurrenceSequence(1,1,1,7)
print([A022097(n) for n in range(41)]) # G. C. Greubel, Jun 03 2025

a(n) = a(n-1) + a(n-2) for n>=2, a(0)=1, a(1)=7, a(-1):=6.
G.f.: (1+6*x)/(1-x-x^2).
a(n) = A109754(6, n+1).
a(n) = A118654(3, n).
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(-13 + sqrt(5)) + (1 + sqrt(5))^n*(13 + sqrt(5))))/sqrt(5). - Herbert Kociemba
a(n) = 6*A000045(n) + A000045(n+1). - R. J. Mathar, Aug 10 2012
a(n) = 8*A000045(n) - A000045(n-2). - Bruno Berselli, Feb 20 2017
From Aamen Muharram, Aug 05 2022: (Start)
a(n) = F(n-4) + F(n-1) + F(n+4),
a(n) = F(n) + F(n+4) - F(n-3),
where F(n) = A000045(n) is the Fibonacci numbers. (End)

A007584 9-gonal (or enneagonal) pyramidal numbers: a(n) = n(n+1)(7*n-4)/6.

Original entry on oeis.org

0, 1, 10, 34, 80, 155, 266, 420, 624, 885, 1210, 1606, 2080, 2639, 3290, 4040, 4896, 5865, 6954, 8170, 9520, 11011, 12650, 14444, 16400, 18525, 20826, 23310, 25984, 28855, 31930, 35216, 38720, 42449, 46410, 50610, 55056, 59755, 64714, 69940, 75440, 81221

Offset: 0

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

For n > 1, the digital roots of this sequence A010888(A007584(n)) form the purely periodic 27-cycle 1, 1, 7, 8, 2, 5, 6, 3, 3, 4, 4, 1, 2, 5, 8, 9, 6, 6, 7, 7, 4, 5, 8, 2, 3, 9, 9. For n > 1, the units digits of this sequence A010879(A007584(n)) form the purely periodic 20-cycle 1, 0, 4, 0, 5, 6, 0, 4, 5, 0, 6, 0, 9, 0, 0, 6, 5, 4, 0, 0. - Ant King, Oct 30 2012

Partial sums of A001106. - Joerg Arndt, Jun 10 2013

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A093564 ((7, 1) Pascal, column m=3).

Cf. similar sequences listed in A237616.

I:=[0, 1, 10, 34, 80]; [n le 5 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 10 2013

a:=n->sum((n+j)^2-(n+j), j=0..n): seq(a(n)/2, n=0..30); # Zerinvary Lajos, May 26 2008

Table[n*(n+1)(7n-4)/6, {n, 0,100}] (* Vladimir Joseph Stephan Orlovsky, Jun 25 2009 *)
LinearRecurrence[{4,-6,4,-1},{1,10,34,80},30] (* Ant King, Oct 27 2012 *)
CoefficientList[Series[x (1 + 6 x) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 10 2013 *)

A007584[n]:=n*(n+1)*(7*n-4)/6$
makelist(A007584[n],n,0,30); /* Martin Ettl, Oct 29 2012 */

a(n) = n*(n+1)*(7*n-4)/6; \\ Michel Marcus, Mar 04 2014

a(n) = (7*n-4)*binomial(n+1, 2)/3.
G.f.: x*(1+6*x)/(1-x)^4.
From Ant King, Oct 27 2012: (Start)
a(n) = a(n-1) + n*(7*n-5)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 7.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = (n+1)*(2*A001106(n)+n)/6.
a(n) = A000292(n) + 6*A000292(n-1).
a(n) = A002414(n) + A000292(n-1).
a(n) = A000217(n) + 7*A000292(n-1).
a(n) = binomial(n+2,3) + 6*binomial(n+1,3). (End)
a(n) = Sum_{i = 0..n-1} (n-i)*(7*i+1) for n>0. - Bruno Berselli, Feb 10 2014
a(n) = A080851(7,n-1). - R. J. Mathar, Jul 28 2016
E.g.f.: (x/6)*(6 + 24*x + 7*x^2)*exp(x). - G. C. Greubel, Oct 29 2017

A093563 (6,1)-Pascal triangle.

Original entry on oeis.org

1, 6, 1, 6, 7, 1, 6, 13, 8, 1, 6, 19, 21, 9, 1, 6, 25, 40, 30, 10, 1, 6, 31, 65, 70, 40, 11, 1, 6, 37, 96, 135, 110, 51, 12, 1, 6, 43, 133, 231, 245, 161, 63, 13, 1, 6, 49, 176, 364, 476, 406, 224, 76, 14, 1, 6, 55, 225, 540, 840, 882, 630, 300, 90, 15, 1, 6, 61, 280, 765, 1380

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(6;n,m) gives in the columns m >= 1 the figurate numbers based on A016921, including the octagonal numbers A000567, (see the W. Lang link).
This is the sixth member, d=6, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-2, for d=1..5.
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+5*z)/(1-(1+x)*z).
The SW-NE diagonals give A022096(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins
  1;
  6,  1;
  6,  7,  1;
  6, 13,  8,  1;
  6, 19, 21,  9,  1;
  6, 25, 40, 30, 10,  1;
  ...

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.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Wolfdieter Lang, First 10 rows and array of figurate numbers

Row sums: A005009(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 5 for n=2 and 0 else.
Cf. A007318, A093564 (d=7), A228196, A228576.
The column sequences give for m=1..9: A016921, A000567 (octagonal), A002414, A002419, A051843, A027810, A034265, A054487, A055848.

a093563 n k = a093563_tabl !! n !! k
a093563_row n = a093563_tabl !! n
a093563_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [6, 1]
-- Reinhard Zumkeller, Aug 31 2014

lim = 11; s = Series[(1 + 5*x)/(1 - x)^(m + 1), {x, 0, lim}]; t = Table[ CoefficientList[s, x], {m, 0, lim}]; Flatten[ Table[t[[j - k + 1, k]], {j, lim + 1}, {k, j, 1, -1}]] (* Jean-François Alcover, Sep 16 2011, after g.f. *)

from math import comb, isqrt
def A093563(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+5*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m)=F(6;n-m, m) for 0<= m <= n, otherwise 0, with F(6;0, 0)=1, F(6;n, 0)=6 if n>=1 and F(6;n, m):= (6*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)=6 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+5*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 5*C(n-1, k). - Philippe Deléham, Aug 28 2005
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(6 + 13*x + 8*x^2/2! + x^3/3!) = 6 + 19*x + 40*x^2/2! + 70*x^3/3! + 110*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

A027818 a(n) = (n+1)*binomial(n+6,6).

Original entry on oeis.org

1, 14, 84, 336, 1050, 2772, 6468, 13728, 27027, 50050, 88088, 148512, 241332, 379848, 581400, 868224, 1268421, 1817046, 2557324, 3542000, 4834830, 6512220, 8665020, 11400480, 14844375, 19143306, 24467184, 31011904, 39002216, 48694800, 60381552, 74393088, 91102473

Offset: 0

Thi Ngoc Dinh (via R. K. Guy)

Number of 13-subsequences of [ 1, n ] with just 6 contiguous pairs.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A093564 ((7, 1) Pascal, column m=7). Partial sums of A050403.

Cf. A062190, A007318, A000142, A245334.

List([0..30], n-> (n+1)*Binomial(n+6,6)); # G. C. Greubel, Aug 29 2019

a027818 n = (n + 1) * a007318' (n + 6) 6
-- Reinhard Zumkeller, Aug 31 2014

[(n+1)*Binomial(n+6,6): n in [0..30]]; // G. C. Greubel, Aug 29 2019

seq((n+1)*binomial(n+6,6), n=0..30); # Zerinvary Lajos, Oct 19 2006

Table[(n+1)*Binomial[n+6,6], {n,0,30}] (* G. C. Greubel, Aug 29 2019 *)

a(n) = (n+1)*binomial(n+6,6) \\ Charles R Greathouse IV, Jun 11 2015

[(n+1)*binomial(n+6,6) for n in (0..30)] # G. C. Greubel, Aug 29 2019

G.f.: (1+6*x)/(1-x)^8.
a(n) = A245334(n+6,6)/A000142(6). - Reinhard Zumkeller, Aug 31 2014
E.g.f.: (7! +9360*x +20520*x^2 +15000*x^3 +4650*x^4 +666*x^5 +43*x^6 + x^7)*exp(x)/7!. - G. C. Greubel, Aug 29 2019
From Amiram Eldar, Jan 28 2022: (Start)
Sum_{n>=0} 1/a(n) = Pi^2 - 5269/600.
Sum_{n>=0} (-1)^n/a(n) = Pi^2/2 - 512*log(2)/5 + 40189/600. (End)

A051740 Partial sums of A007584.

Original entry on oeis.org

1, 11, 45, 125, 280, 546, 966, 1590, 2475, 3685, 5291, 7371, 10010, 13300, 17340, 22236, 28101, 35055, 43225, 52745, 63756, 76406, 90850, 107250, 125775, 146601, 169911, 195895, 224750, 256680, 291896, 330616, 373065, 419475, 470085, 525141

Offset: 0

Barry E. Williams, Dec 07 1999

Convolution of A000027 with A001106 (excluding 0). - Bruno Berselli, Dec 07 2012

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Murray R.Spiegel, Calculus of Finite Differences and Difference Equations, "Schaum's Outline Series", McGraw-Hill, 1971, pp. 10-20, 79-94.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A001106, A007584.
Cf. A093564 ((7, 1) Pascal, column m=4).
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.

List([0..40], n-> (7*n+4)*Binomial(n+3,3)/4); # G. C. Greubel, Aug 29 2019

/* A000027 convolved with A001106 (excluding 0): */ A001106:=func; [&+[(n-i+1)*A001106(i): i in [1..n]]: n in [1..36]]; // Bruno Berselli, Dec 07 2012

seq((7*n+4)*binomial(n+3,3)/4, n=0..40); # G. C. Greubel, Aug 29 2019

Table[(7*n+4)*Binomial[n+3,3]/4, {n,0,40}] (* G. C. Greubel, Aug 29 2019 *)
LinearRecurrence[{5,-10,10,-5,1},{1,11,45,125,280},40] (* Harvey P. Dale, May 18 2023 *)

vector(40, n, (7*n-3)*binomial(n+2,3)/4) \\ G. C. Greubel, Aug 29 2019

[(7*n+4)*binomial(n+3,3)/4 for n in (0..40)] # G. C. Greubel, Aug 29 2019

a(n) = binomial(n+3, 3)*(7*n+4)/4.
a(n) = (7*n+4)*binomial(n+3, 3)/4.
G.f.: (1+6*x)/(1-x)^5.
a(n) = A080852(7,n). - R. J. Mathar, Jul 28 2016
E.g.f.: (4! + 240*x + 288*x^2 + 88*x^3 + 7*x^4)*exp(x)/4!. - G. C. Greubel, Aug 29 2019

A051877 Partial sums of A051740.

Original entry on oeis.org

1, 12, 57, 182, 462, 1008, 1974, 3564, 6039, 9724, 15015, 22386, 32396, 45696, 63036, 85272, 113373, 148428, 191653, 244398, 308154, 384560, 475410, 582660, 708435, 855036, 1024947, 1220842, 1445592, 1702272, 1994168, 2324784, 2697849, 3117324, 3587409

Offset: 0

Barry E. Williams, Dec 14 1999

Convolution of triangular numbers (A000217) and enneagonal numbers (A001106). - Bruno Berselli, Jul 21 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-16.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A051740; A000217, A001106.

Cf. A093564 ((7, 1) Pascal, column m=5).

List([0..40], n-> (7*n+5)*Binomial(n+4,4)/5); # G. C. Greubel, Aug 29 2019

[(n+1)*(n+2)*(n+3)*(n+4)*(7*n+5)/120 : n in [0..40]]; // Wesley Ivan Hurt, May 02 2015

A051877:=n->binomial(n+4,4)*(7*n+5)/5: seq(A051877(n), n=0..40); # Wesley Ivan Hurt, May 02 2015

Table[(n+1)(n+2)(n+3)(n+4)(7n+5)/120, {n, 0, 40}] (* Vincenzo Librandi, May 03 2015 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,12,57,182,462,1008},40] (* Harvey P. Dale, May 05 2022 *)

vector(40, n, (7*n-2)*binomial(n+3,4)/5) \\ G. C. Greubel, Aug 29 2019

[(7*n+5)*binomial(n+4,4)/5 for n in (0..40)] # G. C. Greubel, Aug 29 2019

a(n) = C(n+4, 4)*(7*n+5)/5.
G.f.: (1+6*x)/(1-x)^6.
From Wesley Ivan Hurt, May 02 2015: (Start)
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)*(7*n+5)/120. (End)
E.g.f.: (5! +1320*x +2040*x^2 +920*x^3 +145*x^4 +7*x^5)*exp(x)/5!

cp's OEIS Frontend

A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n*(7*n-5)/2.

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A029653 Numbers in (2,1)-Pascal triangle (by row).

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A016993 a(n) = 7*n + 1.

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A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

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A022097 Fibonacci sequence beginning 1, 7.

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A007584 9-gonal (or enneagonal) pyramidal numbers: a(n) = n*(n+1)*(7*n-4)/6.

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A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n(7n-5)/2.

A007584 9-gonal (or enneagonal) pyramidal numbers: a(n) = n(n+1)(7*n-4)/6.