A349736 Binomial coefficients C(m,k) such that C(m,k), C(m,k+1), C(m,k+2) with 0 <= k <= m-2 form an increasing arithmetic progression.

7, 1001, 490314, 927983760, 6973199770790, 209769429934732479, 25331521183260952835630, 12289694242827235919344118592, 23955991473971122736214778043009679, 187581456720371323313917970237305876898550, 5898404991626652623457605084827693331568853294440, 744569299056744628602691379013860201165514803170616390880

Offset

1,1

Comments

Exercise A1 of 33rd Putnam Exam in 1972 asked one to prove that there are no four consecutive binomial coefficients C(m,k), C(m,k+1), C(m,k+2), C(m,k+3) in arithmetic progression (see link and reference).

However, as there exist such progressions with 3 terms, this sequence lists the smallest term of these arithmetic progressions.

Three consecutive binomial coefficients form an arithmetic progression iff m = n^2+4n+2, n >= 1 (2nd comment of A008865), and then, corresponding k = (n^2+3n-2)/2. Successive pairs (m,k) are (7,1), (14,4), (23,8), (34,13), (47,19), ...

By symmetry of Pascal's triangle with C(m,k) = C(m,m-k), there exists another decreasing arithmetic progression with the same 3 terms in the same row.

Corresponding common differences are in A349737.

References

G. L. Alexanderson, L. F. Klosinski and L. C. Larson, The William Lowell Putnam Mathematical Competition, Problems and Solutions 1965-1984, The Mathematical Association of America, 1985, page 17.

Links

Andrew Howroyd, Table of n, a(n) for n = 1..50

John Scholes, 33rd Putnam 1972, Problem A1.

Index to sequences related to Olympiads and other Mathematical competitions.

Formula

a(n) = C(n^2+4n+2,(n^2+3n-2)/2) = C(A008865(n+2),A034856(n)), for n >= 1.

a(n) ~ c*2^(n^2+4*n)/n, where c = 4*sqrt(2/(Pi*e)). - Stefano Spezia, Nov 29 2021

Example

For n = 1, row 7 of Pascal's triangle is 1, 7, 21, 35, 35, 21, 7, 1; C(7,1) = 7, C(7,2) = 21 and C(7,3) = 35 form an arithmetic progression with common difference = 14, hence a(3) = 7 = C(7,1).
For n = 2, row 14 is 1, 14, 91, 364, 1001, 2002, 3003, 3432, 3003, 2002, 1001, 364, 91, 14, 1; C(14,4) = 1001 , C(14,5) = 2002 and C(14,6) = 3003 form an arithmetic progression with common difference = 1001, hence a(4) = 1001 = C(14,4).

Maple

Sequence = seq(binomial(n^2+4*n+2, (n^2+3*n-2)/2), n=1..16);

Mathematica

nterms=15; Table[Binomial[n^2+4n+2, (n^2+3n-2)/2], {n, nterms}] (* Paolo Xausa, Nov 29 2021 *)

Prog

(PARI) a(n) = binomial(n^2+4*n+2, (n^2+3*n-2)/2) \\ Andrew Howroyd, Oct 29 2023

Crossrefs

Cf. A007318, A008865, A034856, A349737.

Sequence in context: A213960 A173852 A062841 * A110718 A293142 A013544

Adjacent sequences: A349733 A349734 A349735 * A349737 A349738 A349739

Keyword

nonn,easy

Author

Bernard Schott, Nov 28 2021

Extensions

Missing a(9) = 23955...79 inserted by Georg Fischer, Oct 29 2023

Status

approved