A003015 -id:A003015 - cp's OEIS frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719, 427859097160, 1120149658760

Offset: 0

N. J. A. Sloane

a(n)/A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005
Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005
a(n+2) is the Hankel transform of A005807 aerated. - Paul Barry, Nov 04 2008
A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009
The Kn4 sums (see A180662 for definition) of the "Races with Ties" triangle A035317 lead to this sequence. - Johannes W. Meijer, Jul 20 2011
Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012
In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013
For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014
[Note on how to calculate: take the two points (a,b) and (c,d) with aa(n) = A067962(n-1) / A067962(n-2), n > 1. - Reinhard Zumkeller, Sep 24 2015
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017

			G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
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).

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
A. Brousseau, A sequence of power formulas, Fib. Quart., 6 (1968), 81-83.
Alfred Brousseau, Fibonacci and Related Number Theoretic Tables, Fibonacci Association, San Jose, CA, 1972. See p. 17.
Ömer Egecioglu, Elif Saygi, and Zülfükar Saygi, The Mostar index of Fibonacci and Lucas cubes, arXiv:2101.04740 [math.CO], 2021. Mentions this sequence.
Shalosh B. Ekhad and Doron Zeilberger, Automatic Counting of Tilings of Skinny Plane Regions, arXiv preprint arXiv:1206.4864 [math.CO], 2012.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Dale Gerdemann, Golden Ratio Base Digit Patterns for Columns of the Fibonomial Triangle, "Another interesting pattern is for Golden Rectangle Numbers A001654. I made a short video illustrating this pattern, along with other columns of the Fibonomial Triangle A010048".
Jonny Griffiths and Martin Griffiths, Fibonacci-related sequences via iterated QRT maps, Fib. Q., 51 (2013), 218-227.
James P. Jones and Péter Kiss, Representation of integers as terms of a linear recurrence with maximal index, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25. (1998) pp. 21-37. See Lemma 4.1 p. 34.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
C. Pita, On s-Fibonomials, J. Int. Seq. 14 (2011) # 11.3.7.
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
Prapanpong Pongsriiam, Integral Values of the Generating Functions of Fibonacci and Lucas Numbers, College Math. J., 48 (No. 2 2017), pp 97ff.
M. Renault, Dissertation
Wikipedia, Illustration of 273 as a golden rectangle number.
Robert G. Wilson v, Letter to N. J. A. Sloane, circa 1993
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A001655, A001656, A001657, A001658, A010048, A067962.
Cf. A005968, A005969, A098531, A098532, A098533, A119283, A128697.
Cf. A000071, A079472, A080145, A239798, A290565.
Bisection of A006498, A070550, A080239.
First differences of A064831.
Partial sums of A007598.

a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
-- Reinhard Zumkeller, Jun 08 2013

I:=[0,1,2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018

with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
seq(A001654(n), n=0..28); # Zerinvary Lajos, Oct 07 2007

LinearRecurrence[{2,2,-1}, {0,1,2}, 100] (* Vladimir Joseph Stephan Orlovsky, Jul 03 2011 *)
Times@@@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Aug 18 2011 *)
Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)

A001654(n)=fibonacci(n)*fibonacci(n+1);

b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
vector(30, n, b(n-1, 2))  \\ Joerg Arndt, May 08 2016

from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
[a(n) for n in range(101)] # Indranil Ghosh, Aug 03 2017

from math import prod
from gmpy2 import fib2
def A001654(n): return prod(fib2(n+1)) # Chai Wah Wu, May 19 2022

a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2).
a(n) = A006498(2*n-1).
a(n) = a(n - 1) + A007598(n) = a(n - 1) + A000045(n)^2 = Sum_{j <= n} Fibonacci(j)^2. - Henry Bottomley, Feb 09 2001 [corrected by Ridouane Oudra, Apr 12 2025]
For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002.
G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). (Simon Plouffe in his 1992 dissertation; see Comments to A055870),
a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004
a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010
a(n) = (A002878(n) - (-1)^n)/5. - R. J. Mathar, Jul 22 2010
a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010
a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010
Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011
From Tim Monahan, Jul 11 2011: (Start)
a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
From Wolfdieter Lang, Jul 21 2012: (Start)
a(n) = (2*A059840(n+2) - A027941(n))/3, n >= 0, with A059840(n+2) = Sum_{k=0..n} F(k)*F(k+2) and A027941(n) = A001519(n+1) - 1, n >= 0, where A001519(n+1) = F(2*n+1). (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012
a(-1-n) = -a(n) for all n in Z. - Michael Somos, Sep 19 2014
0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
a(n) = (L(2*n+1) - (-1)^n)/5 with L(k) = A000032(k). - J. M. Bergot, Apr 15 2016
E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016
From Klaus Purath, Apr 24 2019: (Start)
a(n) = A061646(n) - Fibonacci(n-1)^2.
a(n) = (A061646(n+1) - A061646(n))/2. (End)
a(n) = A226205(n+1) + (-1)^(n+1). - Flávio V. Fernandes, Apr 23 2020
Sum_{n>=1} 1/a(n) = A290565. - Amiram Eldar, Oct 06 2020
Product_{n>=2} (1 + (-1)^n/a(n)) = phi^2/2 (A239798). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} F(3*k)/F(k) * x^k/k ), where F(n) = A000045(n). - Seiichi Manyama, May 07 2025

Extended by Wolfdieter Lang, Jun 27 2000

A003016 Number of occurrences of n as an entry in rows <= n of Pascal's triangle (A007318).

Original entry on oeis.org

0, 3, 1, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 4, 2, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 0

N. J. A. Sloane

Or, number of occurrences of n as a binomial coefficient. [Except for 1 which occurs infinitely many times. This is the only reason for the restriction "row <= n" in the definition. Any other number can only appear in rows <= n. - M. F. Hasler, Feb 16 2023]
Sequence A138496 gives record values and where they occur. - Reinhard Zumkeller, Mar 20 2008
Singmaster conjectures that this sequence is bounded. - Michael J. Hardy, Jun 09 2025

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 93, #47.
C. S. Ogilvy, Tomorrow's Math. 2nd ed., Oxford Univ. Press, 1972, p. 96.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
H. L. Abbott, P. Erdős and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (1974), 256-261.
Daniel Kane, New Bounds on the Number of Representations of t as a Binomial Coefficient, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 4, Paper A7, 2004.
Kaisa Matomäki, Maksym Radziwill, Xuancheng Shao, Terence Tao, and Joni Teräväinen, Singmaster's conjecture in the interior of Pascal's triangle, arXiv:2106.03335 [math.NT], 2021.
D. Singmaster, How often does an integer occur as a binomial coefficient?, Amer. Math. Monthly, 78 (1971), 385-386.
Eric Weisstein's World of Mathematics, Pascal's Triangle
Index entries for triangles and arrays related to Pascal's triangle

Cf. A003015, A059233, A138496, A180058.

a003016 n = sum $ map (fromEnum . (== n)) $
                      concat $ take (fromInteger n + 1) a007318_tabl
-- Reinhard Zumkeller, Apr 12 2012

a[0] = 0; t = {{1}}; a[n_] := Count[ AppendTo[t, Table[ Binomial[n, k], {k, 0, n}]], n, {2}]; Table[a[n], {n, 0, 101}] (* Jean-François Alcover, Feb 20 2012 *)

{A003016(n)=if(n<4, [0,3,1,2][n+1], my(c=2, k=2, r=sqrtint(2*n)+1, C=r*(r-1)/2); until(, while(C= r\2 && break; C *= r-k; C \= r; r -= 1); c)} \\ M. F. Hasler, Feb 16 2023

from math import isqrt # requires python3.8 or higher
def A003016(n):
    if n < 4: return[0,3,1,2][n]
    cnt = k = 2; r = isqrt(2*n)+1; C = r*(r-1)//2
    while True:
       while C < n and k < r//2:
          C *= r-k; k += 1; C //= k
       if C == n: cnt += 2 - (r == 2*k)
       if k >= r//2: return cnt
       C *= r-k; C //= r; r -= 1 # M. F. Hasler, Feb 16 2023

More terms from Erich Friedman

Edited by N. J. A. Sloane, Nov 18 2007, at the suggestion of Max Alekseyev

A059233 Number of rows in which n appears in Pascal's triangle A007318.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Fabian Rothelius, Jan 20 2001

Central binomial coefficients c = A000984(n) > 1 appear once in the middle column C(2n, n), and thereafter in one or more later rows to the left as C(r,k) and to the right as C(r, r-k), k < r/2; the last time in row r = c = C(c,1) = C(c,c-1). For these, a(n) = (A003016(n)+1)/2. For all other numbers n > 1, a(n) = A003016(n)/2. - M. F. Hasler, Mar 01 2023

			6 appears in both row 4 and row 6 in Pascal's triangle, therefore a(6) = 2.

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 93, #47.
C. S. Ogilvy, Tomorrow's Math. 2nd ed., Oxford Univ. Press, 1972, p. 96.

T. D. Noe, Table of n, a(n) for n=2..10000
D. Singmaster, How often does an integer occur as a binomial coefficient?, Amer. Math. Monthly, 78 (1971), 385-386.
Eric Weisstein's World of Mathematics, Pascal's Triangle
Wikipedia, Singmaster's conjecture
Index entries for triangles and arrays related to Pascal's triangle

Cf. A003016, A003015.

a059233 n = length $ filter (n `elem`) $
                            take (fromInteger n) $ tail a007318_tabl
a059233_list = map a059233 [2..]
-- Reinhard Zumkeller, Dec 24 2012

nmax = 101; A007318 = Table[Binomial[n, k], {n, 0, nmax}, {k, 0, n}]; a[n_] := Position[A007318, n][[All, 1]] // Union // Length; Table[a[n], {n, 2, nmax}] (* Jean-François Alcover, Sep 09 2013 *)

A059233(n)=A003016(n)\/2 \\ M. F. Hasler, Mar 01 2023

a(A180058(n)) = n and a(m) < n for m < A180058(n); a(A182237(n)) = 2; a(A098565(n)) = 3. - Reinhard Zumkeller, Dec 24 2012

a(n) = ceiling(A003016(n)/2). - M. F. Hasler, Mar 01 2023

A098565 Numbers that appear as binomial coefficients exactly 6 times.

Original entry on oeis.org

120, 210, 1540, 7140, 11628, 24310, 61218182743304701891431482520

Offset: 1

Paul D. Hanna, Oct 27 2004

Jean-Marie de Koninck, Nicolas Doyon, and William Verreault, Repetitions of multinomial coefficients and a generalization of Singmaster's conjecture, arXiv:2107.09107 [math.NT], 2021.
Zoe Griffiths, My MegaFavNumber: 61,218,182,743,304,701,891,431,482,520, YouTube video, 2020.
David Singmaster, How Often Does An Integer Occur As A Binomial Coefficient?, American Mathematical Monthly, 78(4), 1971, pp. 385-386; also on Fermat's Library.
Wikipedia, Singmaster's conjecture
Index entries for triangles and arrays related to Pascal's triangle

See A098564 for more information.
Cf. A185024, A182237. Subsequence of A003015.
Cf. A059233.

import Data.List (elemIndices)
a098565 n = a098565_list !! (n-1)
a098565_list = map (+ 2 ) $ elemIndices 3 a059233_list
-- Reinhard Zumkeller, Dec 24 2012

A059233(a(n)) = 3. - Reinhard Zumkeller, Dec 24 2012

a(7) from T. D. Noe, Jul 13 2005

A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).

Original entry on oeis.org

1, 3003, 61218182743304701891431482520

Offset: 1

Eric W. Weisstein, Nov 23 2003 and Wolfdieter Lang, Dec 01 2003

These numbers are known to occur at least six times in Pascal's triangle.
The next term is approximately 3.537 * 10^204 and is in the b-file.
The numbers of digits in a(n), n >= 1, are given in A100022.

Hugo Pfoertner, Table of n, a(n) for n = 1..5
A. I. Shirshov, On the equation C(n, m) = C(n+1, m-1), chapter 10 in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am. Math. Soc., 1999, pp. 83-86
D. Singmaster, Repeated binomial coefficients and Fibonacci numbers, Fibonacci Quarterly, 13 (1975), 295-298.
Eric Weisstein's World of Mathematics, Pascal's Triangle

Subsequence of A003015.

Cf. A081016, A089508, A062527, A206351.

a := proc(n) local a,b,s,p; s:= 1+sqrt(5); p:=16^n;
a := 4-2*p*s^(-4*n-1)+(s+2)*s^(4*n-1)/p:
b := 1+p*((s-2)^(1-4*n)/2-s^(-1-4*n)*(2+s)):
GAMMA(a/5)/(GAMMA(b/5)*GAMMA(1+(a-b)/5)) end:
digits := [1, 4, 29, 205, 1412]: A := n -> round(evalf(a(n),digits[n]+10)):
A(4); # Peter Luschny, Jul 15 2017

Table[Binomial[Fibonacci[2k]Fibonacci[2k+1],Fibonacci[2k-1] Fibonacci[2k]-1], {k,4}] (* Harvey P. Dale, Aug 18 2011 *)

A090162(n)=binomial(fibonacci(2*n+1)*fibonacci(2*n),fibonacci(2*n-1)*fibonacci(2*n)-1) \\ M. F. Hasler, Feb 17 2023

def A090162(n): return C(A000045(2*n+1)*A000045(2*n),A000045(2*n-1)*A000045(2*n)-1) # See A007318 for C(.,.). - M. F. Hasler, Feb 17 2023

a(n) = binomial(A089508(n), A081016(n-1)).
a(n) = binomial(A089508(n)+1, A081016(n-1)-1).
a(n) = Gamma(x)/(Gamma(y)*Gamma(1+x-y)) with x = A206351(n+1) and y = A081016(n-1). - Peter Luschny, Jul 15 2017

A022911 Arrange the nontrivial binomial coefficients C(m,k) (2 <= k <= m/2) in increasing order (not removing duplicates); record the sequence of m's.

Original entry on oeis.org

4, 5, 6, 6, 7, 8, 7, 9, 10, 11, 8, 12, 8, 13, 9, 14, 15, 16, 10, 9, 17, 18, 11, 19, 20, 21, 10, 12, 22, 10, 23, 24, 13, 25, 26, 11, 27, 14, 28, 29, 30, 15, 11, 31, 12, 32, 33, 16, 34, 35, 36, 37, 17, 38, 13, 39, 40, 12, 18, 41, 42, 43, 12, 44, 19, 45, 14, 46, 47, 48

Offset: 1

Clark Kimberling

nonn

In case of duplicates, the m values are listed in decreasing order. Thus a(18)=16 and a(19)=10 corresponding to binomial(16,2)=binomial(10,3)=120. - Robert Israel, Sep 18 2018

Robert Israel, Table of n, a(n) for n = 1..10000
Sean A. Irvine, Java program (github)

Cf. A003015, A006987, A022912, A319382.

N:= 10000: # for binomial(n,k) values <= N
Res:= NULL:
for n from 2 while n*(n-1)/2 <= N do
  for k from 2 to n/2 do
    v:= binomial(n,k);
    if v > N then break fi;
    Res:= Res,[v,n,k]
od od:
Res:= sort([Res],proc(p,q) if p[1]<>q[1] then  p[1]q[2] then p[2]>q[2]
  fi end proc):
map(t -> t[2], Res); # Robert Israel, Sep 18 2018

A319382(n) = binomial(a(n),A022912(n)). - Robert Israel, Sep 18 2018

A022912 Arrange the nontrivial binomial coefficients C(m,k) (2 <= k <= m/2) in increasing order (not removing duplicates); record the sequence of k's.

Original entry on oeis.org

2, 2, 2, 3, 2, 2, 3, 2, 2, 2, 3, 2, 4, 2, 3, 2, 2, 2, 3, 4, 2, 2, 3, 2, 2, 2, 4, 3, 2, 5, 2, 2, 3, 2, 2, 4, 2, 3, 2, 2, 2, 3, 5, 2, 4, 2, 2, 3, 2, 2, 2, 2, 3, 2, 4, 2, 2, 5, 3, 2, 2, 2, 6, 2, 3, 2, 4, 2, 2, 2, 3, 2, 2, 2, 5, 2, 3, 4, 2, 2, 2, 2, 3, 2, 2, 2, 6, 2, 3, 4, 2, 2, 2, 5, 2, 3, 2, 2, 2

Offset: 1

Clark Kimberling

nonn

In case of duplicates, the k values are listed in increasing order. Thus a(18)=2 and a(19)=3 corresponding to binomial(16,2)=binomial(10,3)=120.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A003015, A006987, A022911, A319382.

N:= 10000: # for binomial(n, k) values <= N
Res:= NULL:
for n from 2 while n*(n-1)/2 <= N do
  for k from 2 to n/2 do
    v:= binomial(n, k);
    if v > N then break fi;
    Res:= Res, [v, n, k]
od od:
Res:= sort([Res], proc(p, q) if p[1]<>q[1] then  p[1]q[2] then p[2]>q[2]
fi end proc): map(t -> t[3], Res); # Robert Israel, Sep 18 2018

A319382(n) = binomial(A022911(n),a(n)). - Robert Israel, Sep 18 2018

Corrected by Robert Israel, Sep 18 2018

A064224 Numbers having more than one representation as the product of consecutive integers > 1.

Original entry on oeis.org

120, 210, 720, 5040, 175560, 17297280, 19958400, 259459200, 20274183401472000, 25852016738884976640000, 368406749739154248105984000000

Offset: 1

Jon Perry, Sep 22 2001

Solutions to the equations: x(x+1)...(x+m)=y(y+1)...(y+n) with x>1, y>1.
Some patterns are impossible, e.g. x(x+1)(x+2)(x+3)=y(y+1) has been proved impossible.
The early terms in this sequence each have two representations. Is two the maximum possible? The sequence is infinite: for any n, the number n*(n+1)*...*(n^2+n-1) is in this sequence. The next number of this form is 20274183401472000, which is obtained when n=4. - T. D. Noe, Nov 22 2004
Using an improved algorithm I have performed an exhaustive search up to 2.15 * 10^33 and can confirm the terms shown above are all that exist up to that point. For all k = A002378(n) > 2 we can construct a member of this sequence by equating n(n+1)(n+2)...(k-1) to (n+2)(n+3)...(k-1)k. Also, as demonstrated in my examples below, 5040 is related to 720 as 259459200 is to 210. So we also know that 36055954861352887137197787308347629783163600896000000000 and 6244042313569035223343873483125151604764341428027427022254596874567680000000000000 are terms. - Robert Munafo, Aug 17 2007 [edited by Peter Munn, Aug 20 2023]
MacLeod and Barrodale prove that the equation x(x+1)...(x+m-1) = y(y+1)...(y+n-1) has no solutions x>1 and y>1 for the following pairs of (m,n): (2,4), (2,6), (2,8), (2,12), (4,8), (5,10). They also show that (2,3) has two solutions and (3,6) has one solution. They conjecture that (2,k) has no solution for k>3. [T. D. Noe, Jul 29 2009]

			120 is here because 120 = 2*3*4*5 = 4*5*6.
a(2)=210 because we can write 210=5*6*7 or 14*15. The term a(8) = 259459200 = 5*6*7*8*9*10*11*12*13 = 8*9*10*11*12*13*14*15 is related to 210 by adding the intervening integers (8 through 13) to both products.

H. L. Abbott, P. Erdos and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (March 1974), 256-261.
R. A. MacLeod and I. Barrodale, On equal products of consecutive integers, Canad. Math. Bull., 13 (1970) 255-259. [_T. D. Noe_, Jul 29 2009]
Robert Munafo, Page dealing with this sequence

Cf. A003015 (numbers occurring 5 or more times in Pascal's triangle).
Cf. A002378, A160642.
Subsequence of A045619, A100934.
Cf. A163263 (non-overlapping case). [T. D. Noe, Jul 29 2009]

nn=10^10; t3={}; Do[m=0; p=n; While[m++; p=p(n+m); p<=nn, t3={t3, p}], {n, 2, Sqrt[nn]}]; t3=Sort[Flatten[t3]]; lst={}; Do[If[t3[[i]]==t3[[i+1]], AppendTo[lst, t3[[i]]]], {i, Length[t3]-1}]; Union[lst]

import heapq
def aupton(terms, verbose=False):
    p = 2*3; h = [(p, 2, 3)]; nextcount = 4; alst = []; oldv = None
    while len(alst) < terms:
        (v, s, l) = heapq.heappop(h)
        if v == oldv and v not in alst:
            alst.append(v)
            if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
        if v >= p:
            p *= nextcount
            heapq.heappush(h, (p, 2, nextcount))
            nextcount += 1
        oldv, olds, oldl = v, s, l
        v //= s; s += 1; l += 1; v *= l
        heapq.heappush(h, (v, s, l))
    return alst
print(aupton(8, verbose=True)) # Michael S. Branicky, Jun 24 2021

a(1), a(7) and a(8) from T. D. Noe, Nov 22 2004
a(9) and a(10) from Robert Munafo, Aug 13 2007
a(11) from Robert Munafo, Aug 17 2007
Edited by N. J. A. Sloane, Sep 14 2008 at the suggestion of R. J. Mathar

A180058 Smallest number occurring in exactly n rows of Pascal's triangle.

Original entry on oeis.org

2, 6, 120, 3003

Offset: 1

Reinhard Zumkeller, Dec 24 2012

A059233(a(n)) = n and A059233(m) < n for m < a(n).

			.  n  A180058  referred equal binomial coefficients (A007318)  A059233
.  -  -------  ----------------------------------------------  -------
.  1        2   C (2, 1)                                             1
.  2        6   C (4, 2)   C (6, 1)                                  2
.  3      120   C (10, 3)  C (16, 2)  C (120, 1)                     3
.  4     3003   C (14, 6)  C (15, 5)  C (78, 2)   C (3003, 1)        4 .

Cf. A185024, A182237, A098565, A003015, A003016.

import Data.List (elemIndex)
import Data.Maybe (fromJust)
a180058 = (+ 2) . fromJust . (`elemIndex` a059233_list)

A100934 Numbers having more than one representation as the product of consecutive integers.

Original entry on oeis.org

6, 24, 120, 210, 720, 5040, 40320, 175560, 362880, 3628800, 17297280, 19958400, 39916800, 259459200, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000, 6402373705728000, 20274183401472000, 121645100408832000

Offset: 1

T. D. Noe, Nov 22 2004

nonn

All the factorials occur because we allow products to start with 1. See A064224 for a more restrictive case.

			120 is a term since 120 = 1*2*3*4*5 = 2*3*4*5 = 4*5*6.
210 is a term since 210 = 14*15 = 5*6*7.
Other non-factorial terms are:
  175560    = Product_{i=55..57} i = Product_{i=19..22} i,
  17297280  = Product_{i=63..66} i = Product_{i= 8..14} i,
  19958400  = Product_{i= 5..12} i = Product_{i= 3..11} i,
  259459200 = Product_{i= 8..15} i = Product_{i= 5..13} i,
  20274183401472000 = Product_{i=6..20} i = Product_{i=4..19} i.

Michael S. Branicky, Table of n, a(n) for n = 1..26
H. L. Abbott, P. Erdos and D. Hanson, On the numbers of times an integer occurs as a binomial coefficient, Amer. Math. Monthly, (March 1974), 256-261.

Cf. A064224, A003015 (numbers occurring 5 or more times in Pascal's triangle).

nn=10^10; t3={}; Do[m=0; p=n; While[m++; p=p(n+m); p<=nn, t3={t3, p}], {n, Sqrt[nn]}]; t3=Sort[Flatten[t3]]; lst={}; Do[If[t3[[i]]==t3[[i+1]], AppendTo[lst, t3[[i]]]], {i, Length[t3]-1}]; Union[lst]

import heapq
def aupton(terms, verbose=False):
    p = 1*2; h = [(p, 1, 2)]; nextcount = 3; alst = []; oldv = None
    while len(alst) < terms:
        (v, s, l) = heapq.heappop(h)
        if v == oldv and v not in alst:
            alst.append(v)
            if verbose: print(f"{v}, [= Prod_{{i = {s}..{l}}} i = Prod_{{i = {olds}..{oldl}}} i]")
        if v >= p:
            p *= nextcount
            heapq.heappush(h, (p, 1, nextcount))
            nextcount += 1
        oldv, olds, oldl = v, s, l
        v //= s; s += 1; l += 1; v *= l
        heapq.heappush(h, (v, s, l))
    return alst
print(aupton(20, verbose=True)) # Michael S. Branicky, Jun 24 2021

a(18) and beyond from Michael S. Branicky, Jun 24 2021

Showing 1-10 of 19 results. Next

cp's OEIS Frontend

A001654 Golden rectangle numbers: F(n) * F(n+1), where F(n) = A000045(n) (Fibonacci numbers).

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A003016 Number of occurrences of n as an entry in rows <= n of Pascal's triangle (A007318).

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A059233 Number of rows in which n appears in Pascal's triangle A007318.

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A098565 Numbers that appear as binomial coefficients exactly 6 times.

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A090162 Values of binomial(Fibonacci(2k)*Fibonacci(2k+1),Fibonacci(2k-1)*Fibonacci(2k)-1).

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A022911 Arrange the nontrivial binomial coefficients C(m,k) (2 <= k <= m/2) in increasing order (not removing duplicates); record the sequence of m's.

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A090162 Values of binomial(Fibonacci(2k)Fibonacci(2k+1),Fibonacci(2k-1)Fibonacci(2k)-1).