A093560 -id:A093560 - cp's OEIS frontend

A124459 Square array resulting from the bisection of array A124458. (The other array is A093560.)

2, 3, 2, 3, 5, 2, 3, 8, 7, 2, 3, 11, 15, 9, 2, 3, 14, 26, 24, 11, 2, 3, 17, 40, 50, 35, 13, 2, 3, 20, 57, 90, 85, 48, 15, 2, 3, 23, 77, 147, 175, 133, 63, 17, 2, 3, 26, 100, 224, 322, 308, 196, 80, 19, 2, 3, 29, 126, 324, 546, 630, 504, 276, 99, 21, 2, 3, 32, 155, 450, 870, 1176

Offset: 1

Alford Arnold, Nov 09 2006

Apparently the same as A029618 if the first term is ignored. - R. J. Mathar, Jun 18 2008

			Given the square array
1 2 3 3 3 3 3 3 3 3
1 2 4 5 7 8 10 11 13
1 2 5 7 12 15 22 26
1 2 6 9 18 24 40
1 2 7 11 25 35
1 2 8 13 33 (Table A124458)
1 2 9 15
1 2 10
1 2
1
Omit these odd columns:
1 3 3 3 3 3 3 3 3 3 3
1 4 7 10 13 16 19 22 25 28
1 5 12 22 35 51 70 92 117
1 6 18 40 75 126 196 288
1 7 25 65 140 266 462
1 8 33 98 238 504
1 9 42 140 378
1 10 52 192 (Table A093560)
1 11 63
1 12
1
which yields the square array A124459

Cf. A084215 (antidiagonal sums).

Reppasc := proc(n,k) binomial(n+floor(k/2),n) ; end: A124458 := proc(n,k) add(Reppasc(n,i), i=max(0,k-3)..k-1) ; end: A124459 := proc(n,k) A124458(n,2*k) ; end: for d from 1 to 19 do for k from d to 1 by -1 do n := d-k ; printf("%d,",A124459(n,k)) ; od: od: # R. J. Mathar, Jun 18 2008

More terms from R. J. Mathar, Jun 18 2008

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

Original entry on oeis.org

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803

Offset: 0

N. J. A. Sloane, May 24 1994

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd  b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd  b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3)  = 2*L(n - 2);
L(n) - L(n - 4)  = 5*F(n - 2);
L(n) - L(n - 6)  = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

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Index entries for sequences related to Chebyshev polynomials..
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for two-way infinite sequences.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011

Magma
```
[Lucas(n): n in [0..120]];
```

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

{a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

{a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

{a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022

from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023

[(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025

[lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A002411 Pentagonal pyramidal numbers: a(n) = n^2*(n+1)/2.

Original entry on oeis.org

0, 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, 1183, 1470, 1800, 2176, 2601, 3078, 3610, 4200, 4851, 5566, 6348, 7200, 8125, 9126, 10206, 11368, 12615, 13950, 15376, 16896, 18513, 20230, 22050, 23976, 26011, 28158, 30420, 32800, 35301, 37926, 40678

Offset: 0

N. J. A. Sloane

a(n) = n^2(n+1)/2 is half the number of colorings of three points on a line with n+1 colors. - R. H. Hardin, Feb 23 2002
Sum of n smallest multiples of n. - Amarnath Murthy, Sep 20 2002
a(n) = number of (n+6)-bit binary sequences with exactly 7 1's none of which is isolated. A 1 is isolated if its immediate neighbor(s) are 0. - David Callan, Jul 15 2004
Also as a(n) = (1/6)*(3*n^3+3*n^2), n > 0: structured trigonal prism numbers (cf. A100177 - structured prisms; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
If Y is a 3-subset of an n-set X then, for n >= 5, a(n-4) is the number of 5-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
a(n-1), n >= 2, is the number of ways to have n identical objects in m=2 of altogether n distinguishable boxes (n-2 boxes stay empty). - Wolfdieter Lang, Nov 13 2007
a(n+1) is the convolution of (n+1) and (3n+1). - Paul Barry, Sep 18 2008
The number of 3-character strings from an alphabet of n symbols, if a string and its reversal are considered to be the same.
Partial sums give A001296. - Jonathan Vos Post, Mar 26 2011
a(n-1):=N_1(n), n >= 1, is the number of edges of n planes in generic position in three-dimensional space. See a comment under A000125 for general arrangement. Comment to Arnold's problem 1990-11, see the Arnold reference, p.506. - Wolfdieter Lang, May 27 2011
Partial sums of pentagonal numbers A000326. - Reinhard Zumkeller, Jul 07 2012
From Ant King, Oct 23 2012: (Start)
For n > 0, the digital roots of this sequence A010888(A002411(n)) form the purely periodic 9-cycle {1,6,9,4,3,9,7,9,9}.
For n > 0, the units' digits of this sequence A010879(A002411(n)) form the purely periodic 20-cycle {1,6,8,0,5,6,6,8,5,0,6,6,3,0,0,6,1,8,0,0}.
(End)
a(n) is the number of inequivalent ways to color a path graph having 3 nodes using at most n colors.  Note, here there is no restriction on the color of adjacent nodes as in the above comment by R. H. Hardin (Feb 23 2002).  Also, here the structures are counted up to graph isomorphism, where as in the above comment the "three points on a line" are considered to be embedded in the plane. - Geoffrey Critzer, Mar 20 2013
After 0, row sums of the triangle in A101468. - Bruno Berselli, Feb 10 2014
Latin Square Towers: Take a Latin square of order n, with symbols from 1 to n, and replace each symbol x with a tower of height x. Then the total number of unit cubes used is a(n). - Arun Giridhar, Mar 29 2015
This is the case k = n+4 of b(n,k) = n*((k-2)*n-(k-4))/2, which is the n-th k-gonal number. Therefore, this is the 3rd upper diagonal of the array in A139600. - Luciano Ancora, Apr 11 2015
For n > 0, a(n) is the number of compositions of n+7 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
Also the Wiener index of the n-antiprism graph. - Eric W. Weisstein, Sep 07 2017
For n > 0, a(2n+1) is the number of non-isomorphic 5C_m-snakes, where m = 2n+1 or m = 2n (for n >= 2). A kC_n-snake is a connected graph in which the k >= 2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
For n >= 1, a(n-1) is the number of 0°- and 45°-tilted squares that can be drawn by joining points in an n X n lattice. - Paolo Xausa, Apr 13 2021
a(n) is the number of all possible products of n rolls of a six-sided die. This can be easily seen by the recursive formula a(n) = a(n - 1) + 2 * binomial(n, 2) + binomial(n + 1, 2). - Rafal Walczak, Jun 15 2024
a(n) is the number of all triples consisting of nonnegative integers smaller than n such that the sum of the first two integers is less than n. - Ruediger Jehn, Aug 17 2025

			a(3)=18 because 4 identical balls can be put into m=2 of n=4 distinguishable boxes in binomial(4,2)*(2!/(1!*1!) + 2!/2!) = 6*(2+1) = 18 ways. The m=2 part partitions of 4, namely (1,3) and (2,2), specify the filling of each of the 6 possible two-box choices. - _Wolfdieter Lang_, Nov 13 2007

V. I. Arnold (ed.), Arnold's Problems, Springer, 2004, comments on Problem 1990-11 (p. 75), pp. 503-510. Numbers N_1.
Christian Barrientos, Graceful labelings of cyclic snakes, Ars Combin., Vol. 60 (2001), pp. 85-96.
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/5).
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see Vol. 2, p. 2.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Somaya Barati, Beáta Bényi, Abbas Jafarzadeh and Daniel Yaqubi, Mixed restricted Stirling numbers, arXiv:1812.02955 [math.CO], 2018.
Phyllis Chinn and Silvia Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seq., Vol. 6 (2003), Article 03.2.3.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 17.
Lancelot Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, Vol 19 (2016), Article 16.7.3.
C. Krishnamachaki, The operator (xD)^n, J. Indian Math. Soc., Vol. 15 (1923), pp. 3-4. [Annotated scanned copy]
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber., Vol. 30 (1897), pp. 1917-1926. (Annotated scanned copy)
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, Pentagonal Pyramidal Number.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for two-way infinite sequences.
Index to sequences related to polygonal numbers.
Index to sequences related to pyramidal numbers.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001296, A011379, A015223, A015224, A014799, A014800, A127739, A132118, A139600.
A006002(n) = -a(-1-n).
a(n) = A093560(n+2, 3), (3, 1)-Pascal column.
A row or column of A132191.
Second column of triangle A103371.
Cf. similar sequences listed in A237616.

List([0..45], n->n^2*(n+1)/2); # Muniru A Asiru, Feb 19 2018

a002411 n = n * a000217 n  -- Reinhard Zumkeller, Jul 07 2012

[n^2*(n+1)/2: n in [0..40]]; // Wesley Ivan Hurt, May 25 2014

Maple
```
seq(n^2*(n+1)/2, n=0..40);
```

Table[n^2 (n + 1)/2, {n, 0, 40}]
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 6, 18}, 50] (* Harvey P. Dale, Oct 20 2011 *)
Nest[Accumulate, Range[1, 140, 3], 2] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
CoefficientList[Series[x (1 + 2 x) / (1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jan 08 2016 *)

PARI
```
a(n)=n^2*(n+1)/2
```

concat(0, Vec(x*(1+2*x)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 07 2016

Average of n^2 and n^3.
G.f.: x*(1+2*x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(n) = n*Sum_{k=0..n} (n-k) = n*Sum_{k=0..n} k. - Paul Barry, Jul 21 2003
a(n) = n*A000217(n). - Xavier Acloque, Oct 27 2003
a(n) = (1/2)*Sum_{j=1..n} Sum_{i=1..n} (i+j) = (1/2)*(n^2+n^3) = (1/2)*A011379(n). - Alexander Adamchuk, Apr 13 2006
Row sums of triangle A127739, triangle A132118; and binomial transform of [1, 5, 7, 3, 0, 0, 0, ...] = (1, 6, 18, 40, 75, ...). - Gary W. Adamson, Aug 10 2007
G.f.: x*F(2,3;1;x). - Paul Barry, Sep 18 2008
Sum_{j>=1} 1/a(j) = hypergeom([1, 1, 1], [2, 3], 1) = -2 + 2*zeta(2) = A195055 - 2. - Stephen Crowley, Jun 28 2009
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=0, a(1)=1, a(2)=6, a(3)=18. - Harvey P. Dale, Oct 20 2011
From Ant King, Oct 23 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3.
a(n) = (n+1)*(2*A000326(n)+n)/6 = A000292(n) + 2*A000292(n-1).
a(n) = A000330(n)+A000292(n-1) = A000217(n) + 3*A000292(n-1).
a(n) = binomial(n+2,3) + 2*binomial(n+1,3).
(End)
a(n) = (A000330(n) + A002412(n))/2 = (A000292(n) + A002413(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = (24/(n+3)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+3). - Vladimir Kruchinin, Jun 04 2013
Sum_{n>=1} a(n)/n! = (7/2)*exp(1). - Richard R. Forberg, Jul 15 2013
E.g.f.: x*(2 + 4*x + x^2)*exp(x)/2. - Ilya Gutkovskiy, May 31 2016
From R. J. Mathar, Jul 28 2016: (Start)
a(n) = A057145(n+4,n).
a(n) = A080851(3,n-1). (End)
For n >= 1, a(n) = (Sum_{i=1..n} i^2) + Sum_{i=0..n-1} i^2*((i+n) mod 2). - Paolo Xausa, Apr 13 2021
a(n) = Sum_{k=1..n} GCD(k,n) * LCM(k,n). - Vaclav Kotesovec, May 22 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 2 + Pi^2/6 - 4*log(2). - Amiram Eldar, Jan 03 2022

A001296 4-dimensional pyramidal numbers: a(n) = (3n+1)binomial(n+2, 3)/4. Also Stirling2(n+2, n).

Original entry on oeis.org

0, 1, 7, 25, 65, 140, 266, 462, 750, 1155, 1705, 2431, 3367, 4550, 6020, 7820, 9996, 12597, 15675, 19285, 23485, 28336, 33902, 40250, 47450, 55575, 64701, 74907, 86275, 98890, 112840, 128216, 145112, 163625, 183855, 205905, 229881, 255892, 284050, 314470

Offset: 0

N. J. A. Sloane

Permutations avoiding 12-3 that contain the pattern 31-2 exactly once.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
Partial sums of A002411. - Jonathan Vos Post, Mar 16 2006
If Y is a 3-subset of an n-set X then, for n>=6, a(n-5) is the number of 6-subsets of X having at least two elements in common with Y. - Milan Janjic, Nov 23 2007
Starting with 1 = binomial transform of [1, 6, 12, 10, 3, 0, 0, 0, ...]. Equals row sums of triangle A143037. - Gary W. Adamson, Jul 18 2008
Rephrasing the Perry formula of 2003: a(n) is the sum of all products of all two  numbers less than or equal to n, including the squares. Example: for n=3 the sum of these products is 1*1 + 1*2 + 1*3 + 2*2 + 2*3 + 3*3 = 25. - J. M. Bergot, Jul 16 2011
Half of the partial sums of A011379. [Jolley, Summation of Series, Dover (1961), page 12 eq (66).] - R. J. Mathar, Oct 03 2011
Also the number of (w,x,y,z) with all terms in {1,...,n+1} and w < x >= y > z (see A211795). - Clark Kimberling, May 19 2012
Convolution of A000027 with A000326. - Bruno Berselli, Dec 06 2012
This sequence is related to A000292 by a(n) = n*A000292(n) - Sum_{i=0..n-1} A000292(i) for n>0. - Bruno Berselli, Nov 23 2017
a(n-2) is the maximum number of intersections made from the perpendicular bisectors of all pair combinations of n points. - Ian Tam, Dec 22 2020

			G.f. = x + 7*x^2 + 25*x^3 + 65*x^4 + 140*x^5 + 266*x^6 + 462*x^7 + 750*x^8 + 1155*x^9 + ...

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 835.
A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 195.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 227, #16.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 166, Table 10.4/I/3).
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
P. Aluffi, Degrees of projections of rank loci, arXiv:1408.1702 [math.AG], 2014. ["After compiling the results of many explicit computations, we noticed that many of the numbers d_{n,r,S} appear in the existing literature in contexts far removed from the enumerative geometry of rank conditions; we owe this surprising (to us) observation to perusal of [Slo14]."]
S. Butler and P. Karasik, A note on nested sums, J. Int. Seq. 13 (2010), 10.4.4, page 5.
M. Griffiths, Remodified Bessel Functions via Coincidences and Near Coincidences, Journal of Integer Sequences, Vol. 14 (2011), Article 11.7.1.
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Chanticleer Press, NY, 1950, p. 36.
C. Krishnamachaki, The operator (xD)^n, J. Indian Math. Soc., 15 (1923),3-4. [Annotated scanned copy]
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 29.
T. Mansour, Restricted permutations by patterns of type 2-1, arXiv:math/0202219 [math.CO], 2002.
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, Stirling numbers of the 2nd kind.
Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

a(n)=f(n, 2) where f is given in A034261.
Cf. A000217, A000326, A001297, A001298, A002411, A008277, A008517, A094262.
a(n)= A093560(n+3, 4), (3, 1)-Pascal column.
Cf. A220212 for a list of sequences produced by the convolution of the natural numbers with the k-gonal numbers.
Cf. similar sequences listed in A241765 and A254142.
Cf. A000914.

/* A000027 convolved with A000326: */ A000326:=func; [&+[(n-i+1)*A000326(i): i in [0..n]]: n in [0..40]]; // Bruno Berselli, Dec 06 2012

[(3*n+1)*Binomial(n+2,3)/4: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014

A001296:=-(1+2*z)/(z-1)**5; # Simon Plouffe in his 1992 dissertation for sequence without the leading zero

Table[n*(1+n)*(2+n)*(1+3*n)/24, {n, 0, 100}]
CoefficientList[Series[x (1 + 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Table[StirlingS2[n+2, n], {n, 0, 40}] (* Jean-François Alcover, Jun 24 2015 *)
Table[ListCorrelate[Accumulate[Range[n]],Range[n]],{n,0,40}]//Flatten (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,7,25,65},40] (* Harvey P. Dale, Aug 14 2017 *)

t(n)=n*(n+1)/2
for(i=1,30,print1(","sum(j=1,i,j*t(j))))

{a(n) = n * (n+1) * (n+2) * (3*n+1) / 24}; /* Michael Somos, Sep 04 2017 */

[stirling_number2(n+2,n) for n in range(0,38)] # Zerinvary Lajos, Mar 14 2009

a(n) = n*(1+n)*(2+n)*(1+3*n)/24. - T. D. Noe, Jan 21 2008
G.f.: x*(1+2*x)/(1-x)^5. - Paul Barry, Jul 23 2003
a(n) = Sum_{j=0..n} j*A000217(j). - Jon Perry, Jul 28 2003
E.g.f. with offset -1: exp(x)*(1*(x^2)/2! + 4*(x^3)/3! + 3*(x^4)/4!). For the coefficients [1, 4, 3] see triangle A112493.
E.g.f. x*exp(x)*(24 + 60*x + 28*x^2 + 3*x^3)/24 (above e.g.f. differentiated).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 3. - Kieren MacMillan, Sep 29 2008
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Jaume Oliver Lafont, Nov 23 2008
O.g.f. is D^2(x/(1-x)) = D^3(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
a(n) = A153978(n)/2. - J. M. Bergot, Aug 09 2013
a(n) = A002817(n) + A000292(n-1). - J. M. Bergot, Aug 29 2013; [corrected by Cyril Damamme, Feb 26 2018]
a(n) = A000914(n+1) - 2 * A000330(n+1). - Antal Pinter, Dec 31 2015
a(n) = A080852(3,n-1). - R. J. Mathar, Jul 28 2016
a(n) = 1*(1+2+...+n) + 2*(2+3+...+n) + ... + n*n. For example, a(6) = 266 = 1(1+2+3+4+5+6) + 2*(2+3+4+5+6) + 3*(3+4+5+6) + 4*(4+5+6) + 5*(5+6) + 6*(6).- J. M. Bergot, Apr 20 2017
a(n) = A000914(-2-n) for all n in Z. - Michael Somos, Sep 04 2017
a(n) = A000292(n) + A050534(n+1). - Cyril Damamme, Feb 26 2018
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = (6/5) * (47 - 3*sqrt(3)*Pi - 27*log(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = (6/5) * (16*log(2) + 6*sqrt(3)*Pi - 43). (End)

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

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

A093645 (10,1) Pascal triangle.

Original entry on oeis.org

1, 10, 1, 10, 11, 1, 10, 21, 12, 1, 10, 31, 33, 13, 1, 10, 41, 64, 46, 14, 1, 10, 51, 105, 110, 60, 15, 1, 10, 61, 156, 215, 170, 75, 16, 1, 10, 71, 217, 371, 385, 245, 91, 17, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 10, 91, 369, 876, 1344, 1386, 966, 444, 126, 19, 1

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(10;n,m) gives in the columns m >= 1 the figurate numbers based on A017281, including the 12-gonal numbers A051624 (see the W. Lang link).
This is the tenth member, d=10, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653, A093560-5 and A093644 for d=1..9.
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+9*z)/(1-(1+x)*z).
The SW-NE diagonals give A022100(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k), n >= 1, with n=0 value 9. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

			Triangle begins
   1;
  10,  1;
  10, 11,  1;
  10, 21, 12,  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
W. Lang, First 10 rows and array of figurate numbers .

Row sums: 1 for n=0 and A005015(n-1), n >= 1, alternating row sums are 1 for n=0, 9 for n=2 and 0 otherwise.

The column sequences give for m=1..9: A017281, A051624 (12-gonal), A007587, A051799, A051880, A050406, A052254, A056125, A093646.

a093645 n k = a093645_tabl !! n !! k
a093645_row n = a093645_tabl !! n
a093645_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [10, 1]
-- Reinhard Zumkeller, Aug 31 2014

t[0, 0] = 1; t[n_, k_] := Binomial[n, k] + 9*Binomial[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013, after Philippe Deléham *)

a(n, m) = F(10;n-m, m) for 0 <= m <= n, else 0, with F(10;0, 0)=1, F(10;n, 0)=10 if n >= 1 and F(10;n, m):=(10*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)=10 if n >= 1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (1+9*x)/(1-x)^(m+1), m >= 0.
T(n, k) = C(n, k) + 9*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)*(10 + 21*x + 12*x^2/2! + x^3/3!) = 10 + 31*x + 64*x^2/2! + 110*x^3/3! + 170*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

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

Wolfdieter Lang, Apr 22 2004

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

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

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

Join[{1},Table[Binomial[n,k]+8Binomial[n-1,k],{n,20},{k,0,n}]//Flatten] (* Harvey P. Dale, Aug 17 2024 *)

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

A093561 (4,1) Pascal triangle.

Original entry on oeis.org

1, 4, 1, 4, 5, 1, 4, 9, 6, 1, 4, 13, 15, 7, 1, 4, 17, 28, 22, 8, 1, 4, 21, 45, 50, 30, 9, 1, 4, 25, 66, 95, 80, 39, 10, 1, 4, 29, 91, 161, 175, 119, 49, 11, 1, 4, 33, 120, 252, 336, 294, 168, 60, 12, 1, 4, 37, 153, 372, 588, 630, 462, 228, 72, 13, 1, 4, 41, 190, 525, 960, 1218

Offset: 0

Wolfdieter Lang, Apr 22 2004

The array F(4;n,m) gives in the columns m >= 1 the figurate numbers based on A016813, including the hexagonal numbers A000384 (see the W. Lang link).
This is the fourth member, d=4, in the family of triangles of figurate numbers, called (d,1) Pascal triangles: A007318 (Pascal), A029653 and A093560, for d=1..3.
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+3*z)/(1-(1+x)*z).
The SW-NE diagonals give A000285(n-1) = Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k), n >= 1, with n=0 value 3. 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
The n-th row polynomial is (4 + 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, Mar 02 2018

			Triangle begins
  [1];
  [4, 1];
  [4, 5, 1];
  [4, 9, 6, 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
P. Bala, A note on the diagonals of a proper Riordan Array
W. Lang, First 10 rows and array of figurate numbers .

Cf. Row sums: A020714(n-1), n>=1, 1 for n=0, alternating row sums are 1 for n=0, 3 for n=2 and 0 otherwise.
Columns m=1..9: A016813, A000384 (hexagonal), A002412, A002417, A034263, A051947, A050483, A052181, A055843.
Cf. A007318, A093562 (d=5), A228196, A228576.

a093561 n k = a093561_tabl !! n !! k
a093561_row n = a093561_tabl !! n
a093561_tabl = [1] : iterate
               (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [4, 1]
-- Reinhard Zumkeller, Aug 31 2014

from math import comb, isqrt
def A093561(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),a:=n-comb(r+1,2))*(r+3*(r-a))//r if n else 1 # Chai Wah Wu, Nov 12 2024

a(n, m) = F(4;n-m, m) for 0<= m <= n, otherwise 0, with F(4;0, 0)=1, F(4;n, 0)=4 if n>=1 and F(4;n, m) = (4*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)=4 if n>=1; a(n, m)= a(n-1, m) + a(n-1, m-1).
G.f. row m (without leading zeros): (1+3*x)/(1-x)^(m+1), m>=0.
T(n, k) = C(n, k) + 3*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)*(4 + 9*x + 6*x^2/2! + x^3/3!) = 4 + 13*x + 28*x^2/2! + 50*x^3/3! + 80*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

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

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A124459 Square array resulting from the bisection of array A124458. (The other array is A093560.)

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A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

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