OFFSET

1,3

COMMENTS

No other terms found up to A000217(30000). - Michel Marcus, Sep 17 2013

Numbers in which alternate digits are equal have one of two forms: b*(100^n-1)/99 or b*(100^n-1)/99*10+floor(b/10), where 0<=b<=99. These numbers are triangular if 8*b*(100^n-1)/99+1=y^2 and 8*(b*(100^n-1)/99*10+floor(b/10))+1=y^2, respectively. These equations can be reduced to search of integral points on a finite number of the elliptic curves: 8*b*(100^k*x^3-1)/99+1=y^2 and 8*(b*(100^k*x^3-1)/99*10+floor(b/10))+1=y^2, where x=100^floor(n/3) and k=0, 1, or 2. Each of them has a finite number of integral points and hence there is a finite number of terms in this sequence. - Max Alekseyev, Apr 26 2015

I've computed the integral points on the aforementioned elliptic curves and verified that the sequence is complete. - Max Alekseyev, Jul 30 2024

CROSSREFS

KEYWORD

nonn,base,fini,full

AUTHOR

Amarnath Murthy, Nov 24 2002

STATUS

approved