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A231006
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Number of months after which a date can fall on the same day of the week, and the two months can have the same number of days, in the Gregorian calendar.
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3
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0, 3, 6, 8, 9, 14, 15, 17, 18, 20, 22, 23, 26, 29, 31, 32, 34, 35, 37, 38, 40, 41, 43, 46, 49, 50, 52, 54, 55, 57, 58, 60, 63, 64, 66, 67, 68, 69, 72, 75, 76, 77, 78, 80, 81, 84, 86, 87, 89, 90, 92, 94, 95, 98, 101, 103, 104, 106, 107, 109, 110, 112, 113, 114, 115, 117, 118
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OFFSET
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0,2
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COMMENTS
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In the Gregorian calendar, a non-century year is a leap year if and only if it is a multiple of 4 and a century year is a leap year if and only if it is a multiple of 400.
Assuming this fact, this sequence is periodic with a period of 4800.
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LINKS
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EXAMPLE
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3 belongs to this sequence because December 1, 2011 falls on the same day as March 1, 2012 and both December and March have 31 days each.
6 belongs to this sequence because January 1, 2012 falls on the same day as July 1, 2012 and both January and July have 31 days each.
8 belongs to this sequence because May 1, 2013 falls on the same day as January 1, 2014 and both May and January have 31 days each.
9 belongs to this sequence because January 1, 2013 falls on the same day as October 1, 2013 and both January and October have 31 days each.
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PROG
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(PARI) m=[0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5]; n=[31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]; y=vector(4800, i, (m[((i-1)%12)+1]+((5*((i-1)\48)+(((i-1)\12)%4)-((i-1)\1200)+((i-1)\4800)-!((i-1)%48)+!((i-1)%1200)-!((i-1)%4800)-!((i-2)%48)+!((i-2)%1200)-!((i-2)%4800))))%7); x=vector(4800, i, n[((i-1)%12)+1]+!((i-2)%48)-!((i-2)%1200)+!((i-2)%4800)); for(p=0, 4800, for(q=0, 4800, if(y[(q%4800)+1]==y[((q+p)%4800)+1]&&x[(q%4800)+1]==x[((q+p)%4800)+1], print1(p", "); break)))
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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