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A068133
First triangular number with digit sum = n-th triangular number.
2
0, 1, 3, 6, 28, 78, 1596, 5995, 67896, 887778, 15997996, 398988876, 9876799878, 299789989975, 35998897988976, 589598998999878, 78999997699698778, 7987899888859999878, 1998997979958978979995, 539799799988999999688778
OFFSET
0,3
COMMENTS
The sum of the digits of triangular numbers in most cases is a triangular number. Conjecture: For every triangular number T there exist infinitely many triangular numbers with sum of digits = T.
From Jon E. Schoenfield, Jun 29 2010: (Start)
For any positive k < 132, it is true that more than half of the positive triangular numbers from T(1) through T(k) have a triangular digit sum. However, for any k > 132, more than half of the positive triangular numbers from T(1) through T(k) have a nontriangular digit sum. (At k = 132, there are 66 triangular and 66 nontriangular.)
There exist only finitely many triangular numbers whose digit sum is T(0)=0 or T(1)=1: T(0)=0 is, of course, the only one with digit sum 0, and T(1)=1 and T(4)=10 are the only two with digit sum 1. However, for digit sums equal to each of at least the next several triangular numbers, the conjecture can be easily confirmed by observing that, e.g., T(2), T(20), T(200), T(2000), etc., all have digit sum T(2)=3; T(2+1), T(20+1), T(200+1), T(2000+1), etc., all have digit sum T(3)=6; T(20+2), T(200+2), T(2000+2), T(20000+2), etc., all have digit sum T(4)=10; and, similarly, for all sufficiently large values of j, triangular numbers of the form T(2*10^j+m), where m = 3, 9, 23, 34, 132, 368, 1332, 3943, 19388, 88248, 244948, 1788848, 9838483, 19994343, respectively, will have digit sums T(5)=15, T(6)=21, ..., T(18)=171, respectively. (End)
FORMULA
a(n) = A000217(A068134(n)). - Andrew Howroyd, Sep 21 2024
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy, Feb 21 2002
EXTENSIONS
More terms from Larry Reeves (larryr(AT)acm.org), Jun 17 2002
Term a(0) inserted and terms a(18) and a(19) added by Jon E. Schoenfield, Jun 29 2010
STATUS
approved