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A054216
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Numbers m such that m^2 is a concatenation of two consecutive decreasing numbers.
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10
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91, 9079, 9901, 733674, 999001, 88225295, 99990001, 8900869208, 9296908812, 9604060397, 9999900001, 326666333267, 673333666734, 700730927008, 972603739727, 999999000001, 34519562953737, 39737862788838, 49917309624956
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OFFSET
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1,1
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COMMENTS
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Obviously b(n) = 100^n - 10^n + 1 = (91, 9901, 999001, 99990001, ...) is a subsequence. Are { b(2), b(4), b(6), b(8) } the only terms of this sequence that are prime? - M. F. Hasler, Mar 30 2008. Answer: The smallest prime in this sequence that is not of the form b(n) is A054216(155) = 811451682377384625400019885321 [Max Alekseyev, Oct 08 2008]. See A145381 for further prime terms.
Other subsequences are c(n) = ( 10^(6n) - 2*10^(5n) - 10^(3n) - 2*10^n + 1 )/3 (n>=2), d(n) = (33/101)*(100^(404n+71)+1)+10^(404n+71) (n>=0) and e(n) = (33/101)*(100^(404n-71)+1)+10^(404n-71) (n>=1). Primes among these include c(10), c(14) and d(0). - M. F. Hasler, Oct 09 2008
A positive integer m is in this sequence if and only if m^2 == -1 (mod 10^k + 1) where k is the number of decimal digits in m. Note that k cannot be odd, since in this case 11 divides 10^k + 1 while -1 is not a square modulo 11. - Max Alekseyev, Oct 09 2008
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REFERENCES
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Luca, Florian, and Pantelimon Stănică. "Perfect Squares as Concatenation of Consecutive Integers." The American Mathematical Monthly 126.8 (2019): 728-734.
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LINKS
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FORMULA
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EXAMPLE
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'8242' + '8242-1' gives 82428241 which is 9079^2.
Leading zeros are not allowed, which is why c(1)=266327 is not in this sequence although c(1)^2 = 070930 070929.
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PROG
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(PARI) isA054216(n)={ 1==[1, -1]*divrem(n^2, 10^(#Str(n^2)\2)) & #Str(n^2)%2==0 }
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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Several corrections and additions from M. F. Hasler, Oct 09 2008
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STATUS
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approved
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