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A078254
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Triangular numbers in which alternate digits are equal.
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0
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0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 171, 595, 666, 5050, 5151
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OFFSET
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1,3
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COMMENTS
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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
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LINKS
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CROSSREFS
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KEYWORD
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nonn,base,fini,more
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AUTHOR
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STATUS
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approved
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