A095666 -id:A095666 - cp's OEIS frontend

A095667 Fifth column (m=4) of (1,4)-Pascal triangle A095666.

4, 17, 45, 95, 175, 294, 462, 690, 990, 1375, 1859, 2457, 3185, 4060, 5100, 6324, 7752, 9405, 11305, 13475, 15939, 18722, 21850, 25350, 29250, 33579, 38367, 43645, 49445, 55800, 62744, 70312, 78540, 87465, 97125, 107559, 118807, 130910, 143910, 157850

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=7, a(n-7) is the number of 4-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007
In this sequence if we do a forward difference, then the 3rd forward difference when considered as a sequence will be an arithmetic progression with common difference 1. The same way the sequence formed by the 3rd forward difference of A047668 will be an arithmetic progression with common difference 8. [From Gopalakrishnan (gopala498(AT)yahoo.co.in), Jun 05 2010]
Row 4 of the convolution array A213550.  [Clark Kimberling, Jun 20 2012]

Partial sums of A060488.

A095667:=n->(n+16)*binomial(n+3,3)/4; seq(A095667(k), k=0..70); # Wesley Ivan Hurt, Sep 25 2013

s1=s2=s3=s4=0;lst={};Do[a=n+(n+2);s1+=a;s2+=s1;s3+=s2;s4+=s3;AppendTo[lst,s3/2],{n,3,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 04 2009 *)

G.f.: (4-3*x)/(1-x)^5.
a(n) = 4*b(n)-3*b(n-1) = (n+16)*binomial(n+3, 3)/4, with b(n):=binomial(n+4, 4)= A000332(n+4, 4).
a(n) = sum_{k=1..n} ( sum_{i=1..k} i*(n-k+4) ). - Wesley Ivan Hurt, Sep 25 2013

A095668 Sixth column (m=5) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 21, 66, 161, 336, 630, 1092, 1782, 2772, 4147, 6006, 8463, 11648, 15708, 20808, 27132, 34884, 44289, 55594, 69069, 85008, 103730, 125580, 150930, 180180, 213759, 252126, 295771, 345216, 401016, 463760, 534072, 612612, 700077, 797202, 904761

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X, then, for n >= 8, a(n-8) is the number of 5-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

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

Cf. A000389, A095666.

Cf. A000217, A055999.

[(n+20)*Binomial(n+4, 4)/5: n in [0..30]]; // G. C. Greubel, Nov 25 2017

A095668:=n->(n+20)*binomial(n+4, 4)/5: seq(A095668(n), n=0..80); # Wesley Ivan Hurt, Nov 25 2017

Table[(n + 20)*Binomial[n + 4, 4]/5, {n, 0, 50}] (* G. C. Greubel, Nov 25 2017 *)

for(n=0,30, print1((n+20)*binomial(n+4, 4)/5, ", ")) \\ G. C. Greubel, Nov 25 2017

G.f.: (4-3*x)/(1-x)^6.
a(n) = (n+20)*binomial(n+4, 4)/5.
a(n) = 4*b(n) - 3*b(n-1), with b(n) = binomial(n+5, 5) = A000389(n+5, 5).
E.g.f.: (480 + 2040*x + 1680*x^2 + 440*x^3 + 40*x^4 + x^5)*exp(x)/120. - G. C. Greubel, Nov 25 2017
a(n) = Sum_{i=0..n+1} A000217(i)*A055999(n+2-i). - Bruno Berselli, Mar 05 2018

A095671 Ninth column (m=8) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 33, 153, 525, 1485, 3663, 8151, 16731, 32175, 58630, 102102, 171054, 277134, 436050, 668610, 1001946, 1470942, 2119887, 3004375, 4193475, 5772195, 7844265, 10535265, 13996125, 18407025, 23981724, 30972348, 39674668, 50433900

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=11, a(n-11) is the number of 8-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (9, -36, 84, -126, 126, -84, 36, -9, 1).

CoefficientList[Series[(4-3x)/(1-x)^9,{x,0,30}],x] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{4,33,153,525,1485,3663,8151,16731,32175},30] (* Harvey P. Dale, Jul 07 2015 *)

A095671(n):=(n+32)*binomial(n+7, 7)/8$
makelist(A095671(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

G.f.: (4-3*x)/(1-x)^9.
a(n) = 4*b(n)-3*b(n-1) =(n+32)*binomial(n+7, 7)/8, with b(n):=binomial(n+8, 8)=A000581(n+8, 8).
a(0)=4, a(1)=33, a(2)=153, a(3)=525, a(4)=1485, a(5)=3663, a(6)=8151, a(7)=16731, a(8)=32175, a(n)=9*a(n-1)-36*a(n-2)+84*a(n-3)- 126*a(n-4)+ 126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-9). - Harvey P. Dale, Jul 07 2015

A095819 Tenth column (m=9) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 37, 190, 715, 2200, 5863, 14014, 30745, 62920, 121550, 223652, 394706, 671840, 1107890, 1776500, 2778446, 4249388, 6369275, 9373650, 13567125, 19339320, 27183585, 37718850, 51714975, 70122000, 94103724, 125076072, 164750740

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=12, a(n-12) is the number of 9-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

G.f.: (4-3*x)/(1-x)^10.

a(n) = 4*b(n)-3*b(n-1) =(n+36)*binomial(n+8, 8)/9, with b(n):=binomial(n+9, 9)=A000582(n+9, 9).

A095669 Seventh column (m=6) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 25, 91, 252, 588, 1218, 2310, 4092, 6864, 11011, 17017, 25480, 37128, 52836, 73644, 100776, 135660, 179949, 235543, 304612, 389620, 493350, 618930, 769860, 950040, 1163799, 1415925, 1711696, 2056912, 2457928, 2921688, 3455760, 4068372

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=9, a(n-9) is the number of 6-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

G.f.: (4-3*x)/(1-x)^7.

a(n) = 4*b(n)-3*b(n-1) = (n+24)*binomial(n+5, 5)/6, with b(n):=binomial(n+6, 6)= A000579(n+6, 6).

A095670 Eighth column (m=7) of (1,4)-Pascal triangle A095666.

Original entry on oeis.org

4, 29, 120, 372, 960, 2178, 4488, 8580, 15444, 26455, 43472, 68952, 106080, 158916, 232560, 333336, 468996, 648945, 884488, 1189100, 1578720, 2072070, 2691000, 3460860, 4410900, 5574699, 6990624, 8702320, 10759232, 13217160, 16138848

Offset: 0

Wolfdieter Lang, Jun 11 2004

If Y is a 4-subset of an n-set X then, for n>=10, a(n-10) is the number of 7-subsets of X having at most one element in common with Y. - Milan Janjic, Dec 08 2007

G.f.: (4-3*x)/(1-x)^8.

a(n) = 4*b(n)-3*b(n-1) =(n+28)*binomial(n+6, 6)/7, with b(n):=binomial(n+7, 7)= A000580(n+7, 7).

A020714 a(n) = 5 * 2^n.

Original entry on oeis.org

5, 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 655360, 1310720, 2621440, 5242880, 10485760, 20971520, 41943040, 83886080, 167772160, 335544320, 671088640, 1342177280, 2684354560, 5368709120, 10737418240

Offset: 0

David W. Wilson

Same as Pisot sequences E(5,10), L(5,10), P(5,10), T(5,10). See A008776 for definitions of Pisot sequences.
The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
5 times powers of 2. - Omar E. Pol, Dec 16 2008
Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010
With the addition of "2, 3," at the beginning, this sequence gives terms (n + 3) through the first term greater than 2^n, for n odd, of the negabinary Keith sequence for 2^n, thus proving that with the exception of 2 itself, no odd-indexed power of 2 is a negabinary Keith number (see A188381). - Alonso del Arte, Feb 02 2012
Let b(0) = 5 and b(n+1) = the smallest number not in the sequence such that b(n+1) - Product_{i=0..n} b(i) divides b(n+1) - Sum_{i=0..n} b(i). Then b(n+2) = a(n) for n > 0. - Derek Orr, Jan 15 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..238
John Elias, Illustration: Alternating Tetrahedrons of Tetrahedrons
Tanya Khovanova, Recursive Sequences.
Petro Kosobutskyy, Anastasiia Yedyharova, and Taras Slobodzyan, From Newton's binomial and Pascal's triangle to Collatz's problem, Comp. Des. Sys., Theor. Practice (2023) Vol. 5, No. 1, 121-127.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1003.
Everett Sullivan, Linear chord diagrams with long chords, arXiv preprint arXiv:1611.02771 [math.CO], 2016. See Table 1.
Index entries for linear recurrences with constant coefficients, signature (2).

Row sums of (4, 1)-Pascal triangle A093561.
Row sums of (9, 1)-Pascal triangle A093644.
Row sums of (1, 4)-Pascal triangle A095666 (with leading 4).
Cf. A000079, A001045, A008776, A051916, A118416, A173786, A188381.

[5*2^n: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

Table[5*2^n, {n, 0, 31}] (* Vladimir Joseph Stephan Orlovsky, Dec 16 2008 *)
NestList[2#&,5,40] (* Harvey P. Dale, Mar 13 2022 *)

a(n)=5<Charles R Greathouse IV, Sep 24 2015

a(n) = 5*2^n. a(n) = 2*a(n-1).
G.f.: 5/(1-2*x).
If m is a term greater than 5 of this sequence then m = 5*phi(phi(m)). - Farideh Firoozbakht, Aug 16 2005
a(n) = A118416(n+1,3) for n>2. - Reinhard Zumkeller, Apr 27 2006
a(n) = A000079(n)*5. - Omar E. Pol, Dec 16 2008
a(n) = A173786(n+2,n) for n > 1. - Reinhard Zumkeller, Feb 28 2010
a(n) = A001045(n+4) - A001045(n). - Paul Curtz, Nov 08 2012
Sum_{n>=1} 1/a(n) = 2/5. - Amiram Eldar, Oct 28 2020
E.g.f.: 5*exp(2*x). - Stefano Spezia, May 15 2021

A022095 Fibonacci sequence beginning 1, 5.

Original entry on oeis.org

1, 5, 6, 11, 17, 28, 45, 73, 118, 191, 309, 500, 809, 1309, 2118, 3427, 5545, 8972, 14517, 23489, 38006, 61495, 99501, 160996, 260497, 421493, 681990, 1103483, 1785473, 2888956, 4674429, 7563385, 12237814, 19801199, 32039013, 51840212, 83879225, 135719437

Offset: 0

N. J. A. Sloane

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(5; n-1-k, k), n >= 1, with a(-1)=4. These are the sums of the SW-NE diagonals in P(5; n, k), the (5,1) Pascal triangle A093562. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also sums of the SW-NE diagonals in the (1,4)-Pascal triangle A095666.
Row sums of triangle A131776 starting (1, 5, 6, 11, 17, 28, ...). - Gary W. Adamson, Jul 14 2007
In general, for a Fibonacci sequence beginning with 1,b we have:
a(n) = (2^(-1-n)*((1 - sqrt(5))^n*(1 + sqrt(5) - 2b) + (1 + sqrt(5))^n*(-1 + sqrt(5) + 2b)))/sqrt(5). - Herbert Kociemba, Dec 18 2011
Subsequence of primes: 5, 11, 17, 73, 191, 809, 421493, 1103483, ... . - R. J. Mathar, Aug 09 2012
Pisano periods: 1, 3, 8, 6, 20, 24, 16, 12, 24, 60, 10, 24, 28, 48, 40, 24, 36, 24, 9, 60, ... (differs from A001175). - R. J. Mathar, Aug 10 2012

Muniru A Asiru, Table of n, a(n) for n = 0..960
Élis Gardel da Costa Mesquita, Eudes Antonio Costa, Paula M. M. C. Catarino, and Francisco R. V. Alves, Jacobsthal-Mulatu Numbers, Latin Amer. J. Math. (2025) Vol. 4, No. 1, 23-45. See p. 24.
Tanya Khovanova, Recursive Sequences
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012.
José L. Ramírez and Gustavo N. Rubiano, Properties and Generalizations of the Fibonacci Word Fractal, The Mathematica Journal, Vol. 16 (2014).
Matty van Son, Uniqueness conjectures for extended Markov numbers, arXiv:1911.00746 [math.NT], 2019.
Index entries for linear recurrences with constant coefficients, signature (1,1).

Cf. A000032, A000045, A001175, A093562, A095666, A101220, A109754, A131776.

Row n=4 of A109754 (shifted).

List([0..40],n->4*Fibonacci(n)+Fibonacci(n+1)); # Muniru A Asiru, Mar 04 2018

a0:=1; a1:=5; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..40]]; // Bruno Berselli, Feb 12 2013

with(combinat): a:= n-> 4*fibonacci(n)+fibonacci(n+1): seq(a(n), n=0..32); # Zerinvary Lajos, Oct 05 2007

f[n_] := (LucasL[n - 2] + 8*LucasL[n - 1] + 4*LucasL[n] + 2*LucasL[n + 1])/5; Array[f, 38, 0] (* or *)
LinearRecurrence[{1, 1}, {1, 5}, 38] (* Robert G. Wilson v, Oct 22 2012 *)

a(n)=fibonacci(n-1)+5*fibonacci(n) \\ Charles R Greathouse IV, Jun 05 2011

A022095=BinaryRecurrenceSequence(1,1,1,5)
[A022095(n) for n in range(41)] # G. C. Greubel, Jun 02 2025

a(n) = a(n-1) + a(n-2), n >= 2, a(0)=1, a(1)=5.
G.f.: (1+4*x)/(1-x-x^2).
a(n) = 4*Fibonacci(n) + Fibonacci(n+1), n >= 1. - Zerinvary Lajos, Oct 05 2007, corrected by R. J. Mathar, Apr 07 2011
a(n-1) = ((1 + sqrt(5))^n - (1 - sqrt(5))^n)/(2^n*sqrt(5)) + 2*((1 + sqrt(5))^(n-1) - (1 - sqrt(5))^(n-1))/(2^(n-2)*sqrt(5)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 14 2009
a(n) = 4*Fibonacci(n+2) - 3*Fibonacci(n+1). - Gary Detlefs, Dec 21 2010
a(n) = (L(n-2) + 8*L(n-1) + 4*L(n) + 2*L(n+1))/5 for the Lucas numbers L(n). - J. M. Bergot, Oct 22 2012
a(n) = ((2*sqrt(5) - 1)*(((1 + sqrt(5))/2)^(n+1)) + (2*sqrt(5) + 1)*(((1 - sqrt(5))/2)^(n+1)))/(sqrt(5)). - Bogart B. Strauss, Jul 19 2013
a(n) = Lucas(n-1) + Fibonacci(n+3) = Lucas(n+2) - Fibonacci(n-3). - Greg Dresden and Griffin Donaldson, Mar 03 2022

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

A055999 a(n) = n*(n + 7)/2.

Original entry on oeis.org

0, 4, 9, 15, 22, 30, 39, 49, 60, 72, 85, 99, 114, 130, 147, 165, 184, 204, 225, 247, 270, 294, 319, 345, 372, 400, 429, 459, 490, 522, 555, 589, 624, 660, 697, 735, 774, 814, 855, 897, 940, 984, 1029, 1075, 1122, 1170, 1219, 1269, 1320, 1372, 1425, 1479

Offset: 0

Barry E. Williams, Jun 16 2000

If X is an n-set and Y a fixed (n-4)-subset of X then a(n-3) is equal to the number of 2-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007

Numbers m >= 0 such that 8m+49 is a square. - Bruce J. Nicholson, Jul 28 2017

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Milan Janjic, Two Enumerative Functions.
Amya Luo, Pattern Avoidance in Nonnesting Permutations, Undergraduate Thesis, Dartmouth College (2024). See p. 3.
Leo Tavares, Diamond illustration.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Equals A000217(n+3) - 6.
Cf. A000096, A055998, A074171, A056000, A001477, A002522, A028563, A126890.
Third column (m=2) of (1, 4)-Pascal triangle A095666.
Cf. A000290.

Table[n*(n + 7)/2, {n, 0, 50}] (* G. C. Greubel, Jul 13 2017 *)

a(n)=n*(n+7)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(4-3*x)/(1-x)^3.
a(n) = A126890(n,3) for n>2. - Reinhard Zumkeller, Dec 30 2006
a(n) = A028563(n)/2. - Zerinvary Lajos, Feb 12 2007
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j), then a(n) = -f(n,n-1,4), for n>=1. - Milan Janjic, Dec 20 2008
a(n) = n + a(n-1) + 3 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
a(n) = Sum_{k=1..n} (k+3). - Gary Detlefs, Aug 10 2010
Sum_{n>=1} 1/a(n) = 363/490. - R. J. Mathar, Jul 14 2012
a(n) = 4n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=4..n+3} i. - Wesley Ivan Hurt, Jun 28 2013
E.g.f.: (1/2)*x*(x+8)*exp(x). - G. C. Greubel, Jul 13 2017
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/7 - 319/1470. - Amiram Eldar, Jan 10 2021
a(n) = A000290(n+1) - A000217(n-2). - Leo Tavares, Jan 28 2023
From Amiram Eldar, Feb 12 2024: (Start)
Product_{n>=1} (1 - 1/a(n)) = 15*cos(sqrt(57)*Pi/2)/(8*Pi).
Product_{n>=1} (1 + 1/a(n)) = -63*cos(sqrt(41)*Pi/2)/(8*Pi). (End)

More terms from James Sellers, Jul 04 2000

cp's OEIS Frontend

A095667 Fifth column (m=4) of (1,4)-Pascal triangle A095666.

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A095668 Sixth column (m=5) of (1,4)-Pascal triangle A095666.

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A095671 Ninth column (m=8) of (1,4)-Pascal triangle A095666.

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A095819 Tenth column (m=9) of (1,4)-Pascal triangle A095666.

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A095669 Seventh column (m=6) of (1,4)-Pascal triangle A095666.

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A095670 Eighth column (m=7) of (1,4)-Pascal triangle A095666.

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A020714 a(n) = 5 * 2^n.

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A022095 Fibonacci sequence beginning 1, 5.

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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.