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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.

%I A349736 #44 Oct 30 2023 03:56:55
%S A349736 7,1001,490314,927983760,6973199770790,209769429934732479,
%T A349736 25331521183260952835630,12289694242827235919344118592,
%U A349736 23955991473971122736214778043009679,187581456720371323313917970237305876898550,5898404991626652623457605084827693331568853294440,744569299056744628602691379013860201165514803170616390880
%N 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.
%C A349736 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).
%C A349736 However, as there exist such progressions with 3 terms, this sequence lists the smallest term of these arithmetic progressions.
%C A349736 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), ...
%C A349736 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.
%C A349736 Corresponding common differences are in A349737.
%D A349736 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.
%H A349736 Andrew Howroyd, <a href="/A349736/b349736.txt">Table of n, a(n) for n = 1..50</a>
%H A349736 John Scholes, <a href="https://prase.cz/kalva/putnam/putn72.html">33rd Putnam 1972, Problem A1.</a>
%H A349736 <a href="/index/O#Olympiads">Index to sequences related to Olympiads and other Mathematical competitions</a>.
%F A349736 a(n) = C(n^2+4n+2,(n^2+3n-2)/2) = C(A008865(n+2),A034856(n)), for n >= 1.
%F A349736 a(n) ~ c*2^(n^2+4*n)/n, where c = 4*sqrt(2/(Pi*e)). - _Stefano Spezia_, Nov 29 2021
%e A349736 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).
%e A349736 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).
%p A349736 Sequence = seq(binomial(n^2+4*n+2,(n^2+3*n-2)/2), n=1..16);
%t A349736 nterms=15; Table[Binomial[n^2+4n+2, (n^2+3n-2)/2], {n, nterms}] (* _Paolo Xausa_, Nov 29 2021 *)
%o A349736 (PARI) a(n) = binomial(n^2+4*n+2,(n^2+3*n-2)/2) \\ _Andrew Howroyd_, Oct 29 2023
%Y A349736 Cf. A007318, A008865, A034856, A349737.
%K A349736 nonn,easy
%O A349736 1,1
%A A349736 _Bernard Schott_, Nov 28 2021
%E A349736 Missing a(9) = 23955...79 inserted by _Georg Fischer_, Oct 29 2023